Trigonometric Function Analysis \[ y = 3 \sin 2x \]

Trigonometric Function Analysis: y = 3 sin 2x

Theory

Trigonometric functions are commonly used to describe periodic motion and wave patterns. One of the most important trigonometric functions is the sine function, which produces a smooth, repeating wave. By changing certain parameters, we can adjust the height and speed of this wave.

Concept Explanation

In this lesson, we analyze the function:

\[ y = 3 \sin 2x \]

This function is a transformed version of the basic sine function \(y = \sin x\). The graph oscillates between positive three and negative three, creating a taller and faster wave compared to the standard sine curve.

Some important points on the graph include:

• At \(0^\circ\), the value is \(0\)
• At \(45^\circ\), the value reaches \(3\)
• At \(90^\circ\), the value returns to \(0\)
• At \(135^\circ\), the value reaches \(-3\)
• At \(180^\circ\), the function completes half a cycle

Steps or Formula

The general form of a sine function is:

\[ y = a \sin(kx) \]

From this formula, we can determine the key characteristics:

• Amplitude: \(|a|\)
• Period: \(\dfrac{360^\circ}{k}\)
• Maximum value: \(|a|\)
• Minimum value: \(-|a|\)

For the function \(y = 3 \sin 2x\), we identify:

• \(a = 3\)
• \(k = 2\)

Example Problem

Determine the amplitude and period of the function:

\[ y = 3 \sin 2x \]

Solution:

• Amplitude \(= |3| = 3\)
• Period \(= \dfrac{360^\circ}{2} = 180^\circ\)
• Maximum value \(= 3\)
• Minimum value \(= -3\)

This means the wave completes one full cycle every \(180^\circ\), which is twice as fast as the basic sine function.

Final Answer

The function \(y = 3 \sin 2x\) has an amplitude of 3 and a period of \(180^\circ\). Compared to \(y = \sin x\), it oscillates twice as fast and reaches higher maximum and minimum values. Such functions are useful for modeling real-world oscillating systems, such as mechanical vibrations and mass-spring motions.

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Sine Functions and Waves \[ y = \sin x \]

Sine Functions and Waves

Theory

The sine function is one of the most important functions in trigonometry and is widely used to model periodic phenomena. A periodic function is a function that repeats its values at regular intervals. The sine function creates a smooth and continuous wave pattern, making it ideal for representing cycles and oscillations.

Concept Explanation

The basic sine function is written as:

\[ y = \sin x \]

This graph oscillates between the values 1 and −1. One complete wave cycle starts at zero, rises to a maximum, falls through zero to a minimum, and then returns to zero again.

Key points on the sine graph are:

• At \(0^\circ\), \(y = 0\)
• At \(90^\circ\), \(y = 1\) (maximum)
• At \(180^\circ\), \(y = 0\)
• At \(270^\circ\), \(y = -1\) (minimum)
• At \(360^\circ\), \(y = 0\), completing one full cycle

Steps or Formula

The general form of a sine wave is:

\[ y = a \sin(k(x \pm c)) \]

From this formula, we can determine the main characteristics of the wave:

• Amplitude: \(|a|\)
• Period: \(\dfrac{360^\circ}{k}\)
• Maximum value: \(|a|\)
• Minimum value: \(-|a|\)

The amplitude represents the height of the wave from the center line, while the period shows the horizontal length of one complete cycle.

Example Problem

Consider the function:

\[ y = \sin x \]

Here, the values are:

• \(a = 1\)
• \(k = 1\)

Using the formulas:

• Amplitude \(= |1| = 1\)
• Period \(= \dfrac{360^\circ}{1} = 360^\circ\)
• Maximum value \(= 1\)
• Minimum value \(= -1\)

On the graph, the amplitude can be seen as a vertical distance of 1 from the x-axis to the peak, and the period is the horizontal distance covering one full wave.

Final Answer

The function \(y = \sin x\) produces a regular periodic wave with an amplitude of 1 and a period of \(360^\circ\). Its values range from −1 to 1 and repeat consistently every full cycle. Sine waves are widely used in real-world applications, such as modeling sound waves in physics and audio engineering.

Negative Angle Identities

Negative Angle Identities in Trigonometry

Theory

In trigonometry, angles are not limited to positive values. Angles can also be negative, meaning they are measured clockwise on the unit circle. To work efficiently with negative angles, trigonometry provides special identities known as negative angle identities. These identities help us understand how sine, cosine, and tangent behave when the angle sign changes.

Concept Explanation

Negative angle identities are based on the symmetry of the unit circle. When an angle changes from positive to negative, its point on the unit circle is reflected across the x-axis. Because of this reflection:

• Sine values change sign because sine represents the y-coordinate.
• Cosine values stay the same because cosine represents the x-coordinate.
• Tangent values change sign because tangent is the ratio of sine to cosine.

Steps or Formula

The main negative angle identities are:

• \(\sin(-\theta) = -\sin\theta\)
• \(\cos(-\theta) = \cos\theta\)
• \(\tan(-\theta) = -\tan\theta\)

These formulas allow us to convert negative angles into positive ones, making calculations easier.

Example Problem

Consider the angle \(\theta = 30^\circ\). On the unit circle:

• The point for \(30^\circ\) is \((\cos 30^\circ, \sin 30^\circ)\).
• The point for \(-30^\circ\) is \((\cos(-30^\circ), \sin(-30^\circ))\).

Using the identities:

• \(\sin(-30^\circ) = -\sin 30^\circ = -\frac{1}{2}\)
• \(\cos(-30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}\)
• \(\tan(-30^\circ) = -\tan 30^\circ = -\frac{1}{\sqrt{3}}\)

Now apply the identities to solve this expression:

\[ \sin(-45^\circ) + \cos(-45^\circ) \]

Step by step:

\[ = -\sin 45^\circ + \cos 45^\circ \]

\[ = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \]

Final Answer

\[ 0 \]

Negative angle identities simplify trigonometric problems by using symmetry on the unit circle. By recognizing which functions change sign and which remain the same, calculations become faster and more intuitive.

Unit Circle in Trigonometry $$ x^2 + y^2 = 1 $$

Unit Circle in Trigonometry: Quadrants and Special Angles

In this lesson, we explore the unit circle in trigonometry, with a focus on quadrants and special angles. The unit circle helps us understand the values of sine, cosine, and tangent for different angles in a clear and visual way.

1. Theory of the Unit Circle

The unit circle is a circle with a radius of 1 unit, centered at the origin of the coordinate plane. Every point on the circle satisfies the equation:

$$ x^2 + y^2 = 1 $$

Any point on the unit circle can be written as:

$$ P(\cos \theta, \sin \theta) $$

where θ is the angle measured from the positive x-axis.

2. Concept Explanation

Angles on the unit circle are measured counterclockwise for positive angles and clockwise for negative angles. The circle is divided into four quadrants, and the signs of sine, cosine, and tangent depend on the quadrant where the angle lies.

Important reference points on the unit circle include:

• 0° or 360° → (1, 0)
• 90° → (0, 1)
• 180° → (−1, 0)
• 270° → (0, −1)

3. Steps or Formula

To analyze angles on the unit circle, follow these steps:

1. Determine the quadrant of the angle.
2. Identify the signs of sine, cosine, and tangent in that quadrant.
3. Use special angle values if applicable.

Signs of trigonometric functions in each quadrant:

• Quadrant I: sin +, cos +, tan +
• Quadrant II: sin +, cos −, tan −
• Quadrant III: sin −, cos −, tan +
• Quadrant IV: sin −, cos +, tan −

4. Example Problem

Find the sine, cosine, and tangent values for special angles.

Key trigonometric values on the unit circle:

• 0°: sin = 0, cos = 1, tan = 0
• 30°: sin = \( \frac{1}{2} \), cos = \( \frac{\sqrt{3}}{2} \), tan = \( \frac{1}{\sqrt{3}} \)
• 45°: sin = \( \frac{\sqrt{2}}{2} \), cos = \( \frac{\sqrt{2}}{2} \), tan = 1
• 60°: sin = \( \frac{\sqrt{3}}{2} \), cos = \( \frac{1}{2} \), tan = \( \sqrt{3} \)
• 90°: sin = 1, cos = 0, tan = undefined

As a rotation example, the point (1, 0) rotated 90° counterclockwise becomes (0, 1).

5. Final Answer

The unit circle provides a simple and powerful way to understand trigonometric values. By mastering quadrants and special angles, students can easily determine sine, cosine, and tangent for any angle and build a strong foundation in trigonometry.

Connection Between Arithmetic Series and Trapezoids

Connection Between Arithmetic Series and Trapezoids

In this lesson, we explore a beautiful connection between algebra and geometry. Specifically, we examine how the formula for an arithmetic series is closely related to the formula for the area of a trapezoid.

1. Theory

An arithmetic series is the sum of the terms of an arithmetic sequence. Meanwhile, a trapezoid is a geometric shape with two parallel sides. Although these concepts come from different branches of mathematics, they share a surprisingly similar mathematical structure.

2. Concept Explanation

The key idea lies in the structure of their formulas. Both the arithmetic series formula and the trapezoid area formula use the average of two values, multiplied by a quantity, and divided by two. This similarity reveals how algebraic sums can be interpreted geometrically.

3. Steps or Formula

The formula for the sum of the first n terms of an arithmetic series is:

$$ S_n = \frac{(a_1 + a_n)n}{2} $$

The formula for the area of a trapezoid is:

$$ L = \frac{(s_1 + s_2) \times h}{2} $$

Notice the similarity:

• \(a_1\) and \(a_n\) correspond to the two parallel sides
• \(n\) corresponds to the height
• Both formulas divide the result by 2

4. Example Problem

Find the sum:

$$ 1 + 2 + 3 + \ldots + 99 $$

Step 1: Identify values.

$$ a_1 = 1, \quad a_{99} = 99, \quad n = 99 $$

Step 2: Substitute into the formula.

$$ S_{99} = \frac{(1 + 99) \times 99}{2} $$

Step 3: Simplify.

$$ S_{99} = \frac{100 \times 99}{2} = 4950 $$

5. Final Answer

The sum of the arithmetic series from 1 to 99 is 4950.

This sum can also be visualized geometrically as a set of stepped rectangles that form a trapezoid when rearranged. This elegant connection shows how algebra and geometry work together to reveal deeper mathematical patterns.

Arithmetic Sequences $$ U_{40} = 2 + (40 - 1) \times 3 $$

Arithmetic Sequences

In this lesson, we study arithmetic sequences and learn how to find a specific term in the sequence. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant.

1. Theory of Arithmetic Sequences

An arithmetic sequence follows a regular pattern by adding the same number, called the common difference, to each term. This type of sequence is widely used in mathematics and real-life situations such as calculating linear growth and financial planning.

2. Concept Explanation

To find any term in an arithmetic sequence, we need two key values: the first term and the common difference. Once these are known, we can use a formula to calculate the desired term directly, without listing all previous terms.

3. Steps or Formula

The formula for the nth term of an arithmetic sequence is:

$$ U_n = a + (n - 1)b $$

where:

• a is the first term
• b is the common difference
• n is the term number

4. Example Problem

Find the 40th term of the sequence:

2, 5, 8, 11, ...

Step 1: Identify known values.

$$ a = 2 $$ $$ b = 5 - 2 = 3 $$

Step 2: Substitute into the formula.

$$ U_{40} = 2 + (40 - 1) \times 3 $$

Step 3: Perform the calculation.

$$ U_{40} = 2 + 39 \times 3 $$ $$ U_{40} = 2 + 117 $$ $$ U_{40} = 119 $$

5. Final Answer

The 40th term of the arithmetic sequence is 119.

Arithmetic sequences are useful for modeling steady changes, such as simple interest, regular savings, and evenly increasing patterns. Understanding this concept makes it easier to analyze many real-world problems involving linear growth.

Square Root Approximation $$ \sqrt{99} $$

Square Root Approximation

In this lesson, we learn how to estimate the value of a square root using a simple and effective approximation method. Our focus is to find the approximate value of the square root of 99 without using a calculator.

1. Theory of Square Root Approximation

Square root approximation is a technique used to estimate the value of a square root when the number is not a perfect square. This method is especially useful for quick calculations in mathematics, science, and engineering.

2. Concept Explanation

The key idea is to use the nearest perfect square to the given number. By comparing the number with a nearby perfect square, we can estimate the square root using a simple formula.

3. Steps or Formula

The approximation formula used is:

$$ \sqrt{a} \approx \frac{a + b}{2\sqrt{b}} $$

where:

• a is the number whose square root we want to estimate
• b is the nearest perfect square to a

4. Example Problem

Estimate the value of:

$$ \sqrt{99} $$

Step 1: Identify values.

The nearest perfect square to 99 is 100.

$$ a = 99, \quad b = 100 $$

Step 2: Substitute into the formula.

$$ \sqrt{99} \approx \frac{99 + 100}{2\sqrt{100}} $$

Step 3: Simplify the expression.

$$ \sqrt{100} = 10 $$ $$ \frac{199}{2 \times 10} = \frac{199}{20} $$ $$ = 9.95 $$

5. Final Answer

The approximate value of \(\sqrt{99}\) is 9.95.

This estimate is very close to the actual value of approximately 9.94987, making this method highly accurate. Square root approximation is useful for fast calculations, estimating dimensions, and solving practical problems when exact values are not required.

Infinite Nested Radical Equations $$ \sqrt{2x + \sqrt{2x + \sqrt{2x + \ldots}}} = 20 $$

Infinite Nested Radical Equations

In this lesson, we study an interesting type of equation called an infinite nested radical equation. This equation contains a square root that repeats endlessly, forming a continuous pattern. Our goal is to determine the value of x that satisfies the equation.

1. Theory of Infinite Nested Radicals

An infinite nested radical is an expression where the same square root structure appears again and again. Because the pattern is infinite, we can simplify the problem by representing the entire expression with a single variable.

2. Concept Explanation

The main idea is recognizing that the expression inside the square root is identical to the whole expression. This self-similarity allows us to use substitution, turning a complex-looking equation into a simpler algebraic one.

3. Steps or Formula

1. Let a variable represent the entire infinite radical.
2. Rewrite the equation using the repeating pattern.
3. Substitute the known value into the equation.
4. Square both sides to remove the square root.
5. Solve the resulting equation for x.

4. Example Problem

Given the equation:

$$ \sqrt{2x + \sqrt{2x + \sqrt{2x + \ldots}}} = 20 $$

Step 1: Define the nested expression.

Let the entire infinite radical be represented by y.

$$ y = \sqrt{2x + \sqrt{2x + \sqrt{2x + \ldots}}} $$

From the problem, we know that:

$$ y = 20 $$

Step 2: Use the repeating structure.

Because the pattern repeats infinitely, the expression inside the square root is also y.

$$ y = \sqrt{2x + y} $$

Step 3: Substitute the known value.

$$ 20 = \sqrt{2x + 20} $$

Step 4: Eliminate the square root.

$$ 20^2 = 2x + 20 $$ $$ 400 = 2x + 20 $$

Step 5: Solve for x.

$$ 2x = 380 $$ $$ x = 190 $$

5. Final Answer

The value of x that satisfies the infinite nested radical equation is 190.

This substitution approach is very effective for solving equations with infinite repeating patterns, and it is commonly used in advanced mathematics involving recursive and self-referential expressions.

Infinite Nested Radical Equations $$ \sqrt{2x - \sqrt{2x - \sqrt{2x - \ldots}}} = 20 $$

Infinite Nested Radical Equations

In this lesson, we explore an interesting type of equation known as an infinite nested radical equation. This kind of expression contains a repeating square root pattern that continues endlessly. Our goal is to find the value of x that satisfies the equation.

1. Theory of Infinite Nested Radicals

An infinite nested radical is an expression where the same square root structure repeats forever. Because the pattern never ends, we can take advantage of its self-similar nature by introducing a variable to represent the entire expression.

2. Concept Explanation

The key idea is that the expression inside the square root is identical to the whole expression itself. This allows us to rewrite the equation using substitution, which simplifies the problem into a solvable algebraic equation.

3. Steps or Formula

1. Define a variable to represent the infinite nested expression.
2. Use the repeating pattern to create a new equation.
3. Substitute the known value into the equation.
4. Square both sides to eliminate the square root.
5. Solve the resulting linear equation.

4. Example Problem

Given the equation:

$$ \sqrt{2x - \sqrt{2x - \sqrt{2x - \ldots}}} = 20 $$

Step 1: Define the nested expression.

Let the entire expression be represented by y.

$$ y = \sqrt{2x - \sqrt{2x - \sqrt{2x - \ldots}}} $$

Since the value of the expression is given, we know that:

$$ y = 20 $$

Step 2: Use the repeating pattern.

Because the pattern repeats, the expression inside the square root is also y.

$$ y = \sqrt{2x - y} $$

Step 3: Substitute the known value.

$$ 20 = \sqrt{2x - 20} $$

Step 4: Eliminate the square root.

$$ 20^2 = 2x - 20 $$ $$ 400 = 2x - 20 $$

Step 5: Solve for x.

$$ 2x = 420 $$ $$ x = 210 $$

5. Final Answer

The value of x that satisfies the infinite nested radical equation is 210.

This substitution technique is especially powerful when dealing with equations that contain infinite repeating structures, such as recursive formulas and feedback-based systems.

Systems of Linear Equations $$ x \cdot y = 17 \times 7 = 119 $$

Systems of Linear Equations

In this lesson, we study a system of linear equations involving two variables. The goal is not only to find the values of x and y, but also to determine their product based on the given equations.

1. Theory of Linear Equation Systems

A system of linear equations consists of two or more equations with the same variables. By solving the system, we can find values that satisfy all equations at the same time. Common methods include elimination and substitution.

2. Concept Explanation

We are given two equations. One equation represents the sum of two variables, while the other represents their difference. By combining these equations strategically, we can eliminate one variable and solve for the other.

3. Steps or Formula

1. Add the two equations to eliminate one variable.
2. Solve the resulting equation for the remaining variable.
3. Substitute the known value back into one equation.
4. Find the value of the second variable.
5. Multiply the values of x and y.

4. Example Problem

Given the system of equations:

$$ \begin{cases} x + y = 24 \\ x - y = 10 \end{cases} $$

Step 1: Eliminate one variable.

Add both equations to remove y.

$$ (x + y) + (x - y) = 24 + 10 $$ $$ 2x = 34 $$ $$ x = 17 $$

Step 2: Substitute the value of x.

Substitute x = 17 into the first equation.

$$ 17 + y = 24 $$ $$ y = 7 $$

Step 3: Find the product of x and y.

$$ x \cdot y = 17 \times 7 = 119 $$

5. Final Answer

The product of x and y is 119.

Systems of linear equations are useful in real-life situations such as profit sharing, budget planning, and other problems where totals and differences are known.

Square and Triangle Geometry $$ x = 8 $$

Square and Triangle Geometry

In this lesson, we analyze a geometry problem involving a square and two right triangles. We are given the area of a shaded region and asked to find the value of x, which represents the side length of the square.

1. Theory of Area in Geometry

The area of a triangle can be calculated using the formula one half times the base times the height. By comparing the areas of different shapes, we can form equations that help us find unknown measurements.

2. Concept Explanation

Inside a square with side length x, there are two right triangles. The larger triangle uses the full side of the square as both its base and height. The smaller triangle inside has both its base and height equal to half of the square’s side. The shaded region represents the difference between the areas of these two triangles.

3. Steps or Formula

1. Write the area formula for a triangle.
2. Find the area of the large triangle.
3. Find the area of the small triangle.
4. Subtract the smaller area from the larger area.
5. Set the result equal to the given shaded area.
6. Solve for x.

4. Example Problem

Given:

The shaded green area is 24 square meters. The figure is a square with side length x. Inside the square are a large right triangle and a smaller right triangle.

Step 1: Find the area of the large triangle.

$$ A_{\text{large}} = \frac{1}{2} \cdot x \cdot x = \frac{x^2}{2} $$

Step 2: Find the area of the small triangle.

$$ A_{\text{small}} = \frac{1}{2} \cdot \frac{x}{2} \cdot \frac{x}{2} = \frac{x^2}{8} $$

Step 3: Write the equation for the shaded area.

$$ 24 = \frac{x^2}{2} - \frac{x^2}{8} $$

Step 4: Simplify the equation.

$$ 24 = \frac{4x^2 - x^2}{8} $$ $$ 24 = \frac{3x^2}{8} $$

Step 5: Solve for x.

$$ 3x^2 = 24 \times 8 $$ $$ 3x^2 = 192 $$ $$ x^2 = 64 $$ $$ x = 8 $$

Final Answer:

The value of x, the side length of the square, is 8 meters.

Geometry with Two Circles $$ x = 25 \text{ meters} $$

Geometry with Two Circles

In this lesson, we explore a geometry problem involving two circles placed inside a rectangle. By using coordinate geometry and the distance formula, we can find the unknown length of the rectangle in a clear and systematic way.

1. Theory of Geometry with Circles

When two circles touch each other externally, the distance between their centers is equal to the sum of their radii. By placing the centers of the circles on a coordinate plane, we can use the distance formula to analyze their positions and solve geometric problems accurately.

2. Concept Explanation

In this problem, the rectangle has a fixed height, and two circles of different sizes fit inside it while touching each other. By assigning coordinates to the centers of the circles, we translate the geometric situation into an algebraic equation that can be solved step by step.

3. Steps or Formula

1. Place the centers of the circles on a coordinate plane.
2. Use the fact that touching circles have center distances equal to the sum of their radii.
3. Apply the distance formula.
4. Solve the resulting equation.
5. Add the necessary radius to find the total length of the rectangle.

4. Example Problem

Given:

A rectangle has a height of 18 meters. Inside the rectangle are two circles: a large circle with radius 8 meters and a smaller circle with radius 5 meters.

The center of the large circle is placed at:

$$ (8, 8) $$

The center of the smaller circle is placed at:

$$ (X, 13) $$

Step 1: Use the distance between centers.

Since the circles touch each other, the distance between their centers is:

$$ 8 + 5 = 13 $$

Step 2: Apply the distance formula.

$$ \sqrt{(X - 8)^2 + (13 - 8)^2} = 13 $$

Step 3: Simplify and solve.

$$ (X - 8)^2 + 5^2 = 169 $$ $$ (X - 8)^2 + 25 = 169 $$ $$ (X - 8)^2 = 144 $$ $$ X - 8 = 12 $$ $$ X = 20 $$

Step 4: Find the length of the rectangle.

The total length x is the x-coordinate of the smaller circle plus its radius:

$$ x = 20 + 5 = 25 $$

5. Real-World Application

This type of analytic geometry problem is useful in technical design, layout planning, and space optimization, where precise positioning of circular objects is required.

Final Answer:

$$ x = 25 \text{ meters} $$

Angles in Parallel Lines $$ x + y = 125^\circ $$

Angles in Parallel Lines

Angles formed by parallel lines and a transversal follow special rules that make solving angle problems much easier. In this lesson, we will use these rules to find the value of x + y based on the angles shown in a diagram.

1. Theory of Angles in Parallel Lines

When two parallel lines are cut by a transversal, several types of angles are created. Two important types are corresponding angles and supplementary angles. Corresponding angles are equal in measure, while supplementary angles add up to 180 degrees.

2. Concept Explanation

Corresponding angles appear in the same relative position when a transversal crosses parallel lines, so they have the same measure. Supplementary angles form a straight line, which means their total measure is always 180 degrees. By identifying these relationships, we can solve for unknown angles step by step.

3. Steps or Formula

1. Identify pairs of corresponding angles and set them equal.
2. Solve for the unknown variable.
3. Identify supplementary angles and write an equation.
4. Substitute known values and solve.
5. Add the final values to find x + y.

4. Example Problem

Given:

Two parallel lines are intersected by a transversal. One angle measures 110°, another angle is labeled 2y, and a third angle is labeled x.

Step 1: Use corresponding angles.

Since corresponding angles are equal:

$$ 2y = 110^\circ $$ $$ y = 55^\circ $$

Step 2: Use supplementary angles.

Angles x and 2y form a straight line, so they are supplementary:

$$ x + 2y = 180^\circ $$

Substitute the known value of 2y:

$$ x + 110^\circ = 180^\circ $$ $$ x = 70^\circ $$

Step 3: Find the required sum.

$$ x + y = 70^\circ + 55^\circ = 125^\circ $$

5. Real-World Connection

Understanding corresponding and supplementary angles is essential in real-life applications such as architecture, civil engineering, and navigation, where accurate angle measurement is critical.

Final Answer:

$$ x + y = 125^\circ $$

Determine the domain $$ f(x) = \frac{x + 2}{\sqrt{x^2 - 9}} $$

Domain of a Rational Function with a Square Root

In this lesson, we will learn how to determine the domain of a rational function that contains a square root in the denominator. Finding the domain is an important step because it tells us which values of x make the function valid.

1. Theory of Domain

The domain of a function is the set of all input values for which the function is defined. For rational functions with square roots, special rules apply. We must make sure that the denominator is not zero and that any expression inside a square root is positive.

2. Concept Explanation

When a square root appears in the denominator, the expression inside the square root must be greater than zero. This is more restrictive than simply avoiding zero in the denominator. By solving the resulting inequality, we can determine which values of x are allowed.

3. Steps or Formula

1. Write down the given function.
2. Ensure the denominator is not equal to zero.
3. Require the expression inside the square root to be greater than zero.
4. Solve the resulting inequality.
5. Express the domain using set notation and interval notation.

4. Example Problem

Given function:

$$ f(x) = \frac{x + 2}{\sqrt{x^2 - 9}} $$

Step 1: Apply the domain conditions.

The denominator cannot be zero, and the expression inside the square root must be greater than zero:

$$ x^2 - 9 > 0 $$

Step 2: Solve the inequality.

$$ x^2 - 9 = (x - 3)(x + 3) $$ $$ (x - 3)(x + 3) > 0 $$

The critical points are:

$$ x = -3 \quad \text{and} \quad x = 3 $$

The inequality is satisfied when:

$$ x < -3 \quad \text{or} \quad x > 3 $$

5. Domain Representation

On a number line, the valid values are all numbers to the left of -3 and to the right of 3. These regions represent the domain of the function.

In set notation, the domain is:

$$ D_f = \{x \in \mathbb{R} \mid x < -3 \text{ or } x > 3\} $$

In interval notation, the domain is:

$$ (-\infty, -3) \cup (3, \infty) $$

Final Answer:

The domain of the function consists of all real numbers less than -3 or greater than 3.

Inverse Functions $$ f(2x^2 + x) = 7x - 3 $$

Inverse Functions

Inverse functions help us understand how inputs and outputs of a function can be reversed. In this lesson, we will analyze an inverse function problem step by step and find the value of the inverse function for a given input.

1. Theory of Inverse Functions

An inverse function reverses the effect of the original function. This means that the input and output values are swapped. If a function is written as f(x), its inverse is written as f^-1(x). Graphically, a function and its inverse are reflections of each other across the line y = x.

2. Concept Explanation

When a function is defined using an expression such as f(2x² + x) = 7x - 3, we can find information about its inverse by switching the roles of the input and output. This allows us to express the inverse function in terms of the original variables.

3. Steps or Formula

1. Swap the input and output to represent the inverse relationship.
2. Identify the expression inside the inverse function.
3. Solve for the value that makes the inverse input equal to the given number.
4. Substitute this value back into the inverse expression.

4. Example Problem

Given:

$$ f(2x^2 + x) = 7x - 3 $$

Question:

$$ f^{-1}(4) = ? $$

Step 1: Write the inverse relationship.

Since inverse functions swap inputs and outputs:

$$ f^{-1}(7x - 3) = 2x^2 + x $$

Step 2: Find the value of x that makes the input equal to 4.

$$ 7x - 3 = 4 $$ $$ 7x = 7 $$ $$ x = 1 $$

Step 3: Substitute x into the inverse expression.

$$ f^{-1}(4) = 2(1)^2 + 1 $$ $$ f^{-1}(4) = 2 + 1 = 3 $$

5. Graph Interpretation

The function y = 7x - 3 and its inverse y = \frac{x + 3}{7} are mirror images of each other across the line y = x. This reflection confirms the relationship between a function and its inverse.

Final Answer:

$$ f^{-1}(4) = 3 $$

Quadratic Functions $$ y = x^2 + 2x - 3 $$

Quadratic Functions and Their Graphs

Quadratic functions are an important topic in algebra and are commonly represented by equations in the form of a parabola. In this lesson, we will explore a quadratic function, find its roots, and understand the key features of its graph.

1. Theory of Quadratic Functions

A quadratic function is a polynomial function of degree two. Its general form is y = ax² + bx + c, where a, b, and c are constants and a ≠ 0. The graph of a quadratic function is a parabola, which can open upward or downward depending on the value of a.

2. Concept Explanation

To understand a quadratic function, we often analyze its roots, vertex, and overall shape. The roots are the x-values where the graph crosses the x-axis, found by setting the function equal to zero. The vertex represents either the lowest or highest point of the parabola.

3. Steps or Formula

1. Write the quadratic function.
2. Set the equation equal to zero to find the roots.
3. Factor the quadratic expression.
4. Solve for the x-values.
5. Identify the vertex and key features of the graph.

4. Example Problem

Given function:

$$ y = x^2 + 2x - 3 $$

Step 1: Find the roots by setting the equation to zero.

$$ x^2 + 2x - 3 = 0 $$

Step 2: Factor the quadratic expression.

$$ (x + 3)(x - 1) = 0 $$

Step 3: Solve for x.

$$ x_1 = -3 \quad \text{and} \quad x_2 = 1 $$

These values are the x-intercepts of the graph.

Step 4: Identify the vertex.

The vertex of the parabola is located at:

$$ (-1, -4) $$

This means the lowest point of the graph occurs when y = -4.

5. Graph Interpretation

The graph of the function y = x² + 2x - 3 is a parabola that opens upward. It crosses the x-axis at x = -3 and x = 1, and has its vertex at the point (-1, -4).

Final Answer:

The quadratic function has roots at x = -3 and x = 1, a vertex at (-1, -4), and its graph is an upward-opening parabola.

Adding Square Roots $$ \sqrt{50} + \sqrt{8} $$

Operations with Radicals: Adding Square Roots

In this lesson, we focus on operations with radicals, specifically adding square roots. Before radicals can be added, each square root must be simplified as much as possible. Only like radicals can be combined.

1. Theory of Adding Square Roots

Square roots can only be added together if they have the same radical part. This means the expression inside the square root must be identical. To achieve this, we often need to simplify each radical first.

2. Concept Explanation

Simplifying a square root involves factoring the number inside the radical into a perfect square multiplied by another number. The square root of the perfect square becomes a whole number, while the remaining factor stays inside the radical.

3. Formula or Steps

1. Simplify each square root separately.
2. Check that the radicals are the same.
3. Add the coefficients.
4. Write the final simplified result.

4. Example Problem

Problem:

$$ \sqrt{50} + \sqrt{8} $$

Solution:

Step 1: Simplify each square root.

$$ \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} $$ $$ \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} $$

Step 2: Add the like radical terms.

$$ 5\sqrt{2} + 2\sqrt{2} = (5 + 2)\sqrt{2} $$ $$ = 7\sqrt{2} $$

Final Answer:

$$ \sqrt{50} + \sqrt{8} = 7\sqrt{2} $$

Operations with Fractions $$ \frac{43}{30} = A + \frac{1}{B + \frac{1}{C + \frac{1}{D}}} $$

Operations with Fractions: Continued Fractions

In this lesson, we explore a more advanced operation with fractions by working with continued fractions. The goal is not only to simplify a fraction, but also to identify specific values hidden within its structure.

1. Problem Overview

We are given the following expression:

$$ \frac{43}{30} = A + \frac{1}{B + \frac{1}{C + \frac{1}{D}}} $$

Our task is to determine the values of A, B, C, and D, then find the value of A + B + C + D.

2. Step-by-Step Solution

Step 1: Convert the fraction into a mixed number.

$$ \frac{43}{30} = 1 + \frac{13}{30} $$

This works because a mixed number can be written as the sum of a whole number and a fraction.

Step 2: Convert the fraction into a continued fraction.

Recall the rule:

$$ \frac{a}{b} = \frac{1}{\frac{b}{a}} $$

Applying this rule to the fraction:

$$ \frac{13}{30} = \frac{1}{\frac{30}{13}} $$

Now simplify the denominator step by step:

$$ \frac{30}{13} = 2 + \frac{4}{13} $$ $$ \frac{4}{13} = \frac{1}{\frac{13}{4}} = \frac{1}{3 + \frac{1}{4}} $$

So the continued fraction becomes:

$$ \frac{13}{30} = \frac{1}{2 + \frac{1}{3 + \frac{1}{4}}} $$

3. Identifying the Values

Substituting back into the original expression:

$$ \frac{43}{30} = 1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4}}} $$

From this form, we can identify:

• A = 1
• B = 2
• C = 3
• D = 4

4. Final Calculation

Now add all the values:

$$ A + B + C + D = 1 + 2 + 3 + 4 = 10 $$

Final Answer:

$$ A + B + C + D = 10 $$

Dividing Fractions $$ 5 \div \frac{3}{4} $$

Dividing Fractions with Whole Numbers

Dividing fractions is an important skill in mathematics and follows a clear set of rules. Unlike addition or subtraction, division requires changing the operation before solving. Understanding this process makes fraction division much easier and more meaningful.

1. Theory of Dividing Fractions

When dividing by a fraction, we do not divide directly. Instead, we change the division into multiplication by using the reciprocal of the fraction. The reciprocal is found by swapping the numerator and the denominator.

2. Concept Explanation

The division problem “five divided by three fourths” can be interpreted as asking: “How many groups of three fourths are in five?” This way of thinking helps us understand why multiplication by the reciprocal works.

By counting how many times three fourths fits into five whole units, we can clearly see the meaning behind the final answer.

3. Formula or Steps

1. Change the division into multiplication by the reciprocal.
2. Multiply the numerators and denominators.
3. Simplify the result.
4. Convert to a mixed number if the fraction is improper.

4. Example Problem

Problem:

$$ 5 \div \frac{3}{4} $$

Solution:

Step 1: Change division into multiplication by the reciprocal.

$$ 5 \div \frac{3}{4} = 5 \times \frac{4}{3} $$

Step 2: Multiply the numerators and denominators.

$$ 5 \times \frac{4}{3} = \frac{5 \times 4}{1 \times 3} = \frac{20}{3} $$

Step 3: Simplify the result.

$$ \frac{20}{3} = 6 \frac{2}{3} $$

5. Conceptual Visualization

Conceptually, dividing five by three fourths asks how many groups of three fourths can be formed from five. We can fit six complete groups of three fourths, with two thirds of another group remaining. This confirms the final result.

Final Answer:

$$ 5 \div \frac{3}{4} = \frac{20}{3} = 6 \frac{2}{3} $$

Subtracting Fractions $$ \frac{3}{4} - 2 $$

Subtracting Fractions with Whole Numbers

Subtracting fractions follows similar rules to adding fractions. When a fraction is subtracted from a whole number, both values must be written in the same fractional form. This helps ensure the subtraction is done correctly and avoids confusion.

1. Theory of Subtracting Fractions

In fraction subtraction, a whole number must be converted into a fraction before performing the operation. By giving both numbers the same denominator, we can subtract the numerators directly while keeping the denominator unchanged.

2. Concept Explanation

The key concept in subtracting fractions with whole numbers is understanding equivalent fractions. A whole number can be written as a fraction over 1, then converted into an equivalent fraction that matches the denominator of the given fraction. This allows subtraction to be done easily.

When the result is negative, it means the value is less than zero. Negative fractions can also be written as mixed numbers for better understanding.

3. Formula or Steps

1. Convert the whole number into a fraction.
2. Change the fraction so both denominators are the same.
3. Subtract the numerators and keep the denominator.
4. Simplify the result and write it as a mixed number if needed.

4. Example Problem

Problem:

$$ \frac{3}{4} - 2 $$

Solution:

Step 1: Change the whole number into a fraction.

$$ 2 = \frac{2}{1} $$

Step 2: Make the denominators the same.

$$ \frac{2}{1} = \frac{8}{4} $$

Step 3: Subtract the fractions.

$$ \frac{3}{4} - \frac{8}{4} = \frac{3 - 8}{4} = -\frac{5}{4} $$

Step 4: Simplify the negative result.

$$ -\frac{5}{4} = -1 \frac{1}{4} $$

5. Visual Explanation Using a Number Line

To better understand this subtraction, imagine a number line. Start at three fourths. Then move two whole units to the left, which is the same as moving eight fourths. You will land at negative one and one fourth on the number line.

Final Answer:

$$ \frac{3}{4} - 2 = -\frac{5}{4} = -1 \frac{1}{4} $$

Adding Fractions $$ \frac{3}{4} + 1 = \frac{7}{4} = 1 \frac{3}{4} $$

Adding Fractions with Whole Numbers

Adding fractions is a basic mathematical skill that helps us combine numbers written in different forms. Sometimes, we need to add a fraction and a whole number. To solve this problem correctly, both numbers must be written in the same form before adding them together.

1. Theory of Adding Fractions

When adding fractions and whole numbers, the whole number must first be converted into a fraction. This allows both numbers to have a denominator, making the addition process easier and more accurate.

2. Concept Explanation

The key concept in adding fractions with whole numbers is making sure that both numbers have the same denominator. A whole number can be written as a fraction by placing it over 1. After that, the fraction is adjusted so its denominator matches the other fraction.

3. Formula or Steps

1. Convert the whole number into a fraction.
2. Make the denominators the same.
3. Add the numerators and keep the denominator.
4. Simplify the result if necessary.

4. Example Problem

Problem:

$$ \frac{3}{4} + 1 $$

Solution:

Step 1: Change the whole number into a fraction.

$$ 1 = \frac{1}{1} $$

Step 2: Make the denominators the same.

$$ \frac{1}{1} = \frac{4}{4} $$

Step 3: Add the fractions.

$$ \frac{3}{4} + \frac{4}{4} = \frac{7}{4} $$

Step 4: Simplify the result.

$$ \frac{7}{4} = 1 \frac{3}{4} $$

Final Answer:

$$ \frac{3}{4} + 1 = \frac{7}{4} = 1 \frac{3}{4} $$