Problem 4 - A Graphical Approach to Precalculus with Limits

Question

In Exercises begin by drawing a rough sketch to determine the number of real solutions for the equation \(y_{1}=y_{2}\) Then solve this equation by hand. Give the solution set and any extraneous values that may occur. Do not use a calculator. $$\begin{aligned} &y_{1}=\sqrt{x}\\\ &y_{2}=3 x \end{aligned}$$

Step-by-Step Solution

Answer

The solution set is \( x = 0 \) and \( x = \frac{1}{9} \). No extraneous values.

Step 1: Sketch the Functions

Sketch the graphs of the functions given: \( y_1 = \sqrt{x} \) and \( y_2 = 3x \). The graph of \( y_1 \) is a square root function starting at the origin (0,0) and rising gradually. The graph of \( y_2 \) is a linear function, a straight line with a slope of 3, passing through the origin. By plotting these, we estimate the intersection points.

Step 2: Equate and Solve the Functions

Set the equations equal to each other to find the intersection points: \( \sqrt{x} = 3x \). Square both sides to eliminate the square root, resulting in \( x = 9x^2 \). Rearrange the equation to form \( 9x^2 - x = 0 \).

Step 3: Factor the Equation

Factor out the common term from the equation \( x(9x - 1) = 0 \). Set each factor equal to zero to solve for \( x \). This gives two potential solutions: \( x = 0 \) and \( 9x - 1 = 0 \).

Step 4: Solve the Factor Equation

For the factor \( 9x - 1 = 0 \), solve for \( x \). Adding 1 to both sides gives \( 9x = 1 \), and then divide both sides by 9 to get \( x = \frac{1}{9} \).

Step 5: Verify Solutions

Substitute each solution back into both original equations to ensure they satisfy both. For \( x = 0 \), \( y_1 = 0 \) and \( y_2 = 0 \). For \( x = \frac{1}{9} \), \( y_1 = \sqrt{\frac{1}{9}} = \frac{1}{3} \) equals \( y_2 = 3(\frac{1}{9}) = \frac{1}{3} \). Both solutions are valid; no extraneous values.

Key Concepts

Equations

In mathematics, equations play a crucial role in expressing relationships between different quantities. An equation states that two expressions are equal, often involving variables like x and y. For this exercise, we have the equations

• \( y_1 = \sqrt{x} \)
• \( y_2 = 3x \)

These represent different types of functions: a square root function and a linear function, respectively. The goal is to find the values of x where these two functions equal each other.
To do this mathematically, we set the two equations equal to one another: \( \sqrt{x} = 3x \). This procedure is sometimes referred to as equating the equations. Solving the resulting equation requires eliminating the square root, which we accomplish by squaring both sides, leading us to \( x = 9x^2 \). This transformation is often necessary, especially for functions that include square roots or other more complex expressions.

Graphing Functions

Graphing functions gives a visual perspective on mathematical relationships and solutions. By sketching the graphs for

• \( y = \sqrt{x} \)
• \( y = 3x \)

we can anticipate the intersections, which represent the solutions where the two functions are equal.
The function \( y = \sqrt{x} \) is a basic square root function. It starts at the coordinate (0,0) and curves upward, increasing gradually as x increases. On the other hand, the function \( y = 3x \) is a linear function that also passes through the origin, but it rises steeply with a slope of 3.
By plotting these functions, you visually identify potential points of intersection. The graphical method offers a quick way to derive a rough estimate of solutions, before solving the equations algebraically.

Solution Verification

After finding the values of x where the functions intersect, known as solutions, it's important to verify these solutions. Verification ensures the solutions are accurate and not extraneous. For our solved equation, we found potential solutions at

• \( x = 0 \)
• \( x = \frac{1}{9} \)

We substitute these back into the original functions to see if they satisfy both equations.
- **For \( x = 0 \):**

• \( y_1 = \sqrt{0} = 0 \)
• \( y_2 = 3 \times 0 = 0 \)

Both expressions are equal, confirming the solution.
- **For \( x = \frac{1}{9} \):**

• \( y_1 = \sqrt{\frac{1}{9}} = \frac{1}{3} \)
• \( y_2 = 3 \times \frac{1}{9} = \frac{1}{3} \)

Both function values are again equal, affirming the correctness of this solution as well.
Verification is not just mechanical; it's crucial for identifying whether any steps in solving introduced extraneous solutions.