Problem 49 - Intermediate Algebra

Question

(a) find the inverse of the given function, and (b) graph the given function and its inverse on the same set of axes. (Objective 4) $$f(x)=x^{2}, x \geq 0$$

Step-by-Step Solution

Answer

The inverse is \(f^{-1}(x) = \sqrt{x}\), where \(x \geq 0\). Graph both the function \(f(x) = x^2\) and its inverse, creating a reflection across \(y = x\).

Step 1: Start by Setting the Function Equal to y

To find the inverse of a function, we begin by replacing \(f(x)\) with \(y\). This gives us the equation \(y = x^2\), where \(x \geq 0\).

Step 2: Solve for x in Terms of y

Since we want to express \(x\) in terms of \(y\), take the square root of both sides to get \(x = \sqrt{y}\). Ensure you consider \(x \geq 0\) since the original function only includes non-negative \(x\) values.

Step 3: Express the Inverse Function

In the inverse function, we swap \(x\) and \(y\). Thus, the inverse function is \(f^{-1}(x) = \sqrt{x}\), where \(x \geq 0\).

Step 4: Graph the Function and Its Inverse

To graph \(f(x) = x^2\) and its inverse \(f^{-1}(x) = \sqrt{x}\):1. **Graph \(f(x) = x^2\):** Start graphing from the origin and its curve rises upwards as \(x\) increases (only non-negative \(x\) values included).2. **Graph \(f^{-1}(x) = \sqrt{x}\):** This curve starts at the origin and rises slowly to the right, representing a reflection of \(f(x)\) across the line \(y=x\).3. **Draw the Line \(y = x\):** This is a diagonal line through the origin, showing the mirror symmetry between \(f(x)\) and \(f^{-1}(x)\).

Key Concepts

Graphing Functions

Graphing functions is a fundamental skill in mathematics. It allows us to visualize how input values map to output values. Graphing the function \( f(x) = x^2 \) involves the following steps:

• Start at the origin, which is the point (0, 0). This is because when \( x = 0 \), \( f(x) = 0^2 = 0 \).
• The graph is a parabola that opens upwards. Only non-negative \( x \) values are considered, as \( x \geq 0 \) in our function.
• As \( x \) increases, \( f(x) \) grows rapidly. For example, when \( x = 1, f(x) = 1 \), but when \( x = 2, f(x) = 4 \).

The inverse function of \( f(x) = x^2 \), denoted as \( f^{-1}(x) = \sqrt{x} \), is graphed differently. It starts at the origin and rises slowly to the right. This difference in growth rate is essential to understand the relationship between original and inverse functions.

Square Root Function

The square root function is significant in inverse function problems. For \( f(x) = x^2 \), its inverse is \( f^{-1}(x) = \sqrt{x} \). Here's how it behaves:

• It starts at the origin (0, 0), since \( \sqrt{0} = 0 \).
• The graph rises gradually to the right. Unlike \( x^2 \), \( \sqrt{x} \) indicates a slower increase.
• This function only deals with non-negative inputs \( x \geq 0 \) since square roots of positive numbers are considered.

Understanding the domain and range is critical. The domain of \( f^{-1}(x) = \sqrt{x} \) is \( x \geq 0 \), just like how \( f(x) = x^2 \) is defined for \( x \geq 0 \). Thus, the square root function acts as a reflection of the parabola \( f(x) = x^2 \), but mirrored along the line \( y = x \). This intrinsic reflection characteristic will be discussed further in the next section.

Reflection Across y=x

Reflecting a function's graph across the line \( y = x \) is a common method to verify inverse functions. For a function and its inverse, this reflection should reveal a symmetry.

• When graphing \( f(x) = x^2 \) and \( f^{-1}(x) = \sqrt{x} \), the line \( y = x \) helps illustrate their relationship.
• This line is diagonal through the origin, showing that each point \( (a, b) \) on \( f(x) = x^2 \) corresponds to a point \( (b, a) \) on \( f^{-1}(x) = \sqrt{x} \).
• The graphs are like mirror images across this line, a defining feature of inverse functions.

When you glance at these graphs, you'll notice: \( f(x) = x^2 \) rises sharply compared to the slower increase of \( f^{-1}(x) = \sqrt{x} \). This symmetry demonstrated across \( y = x \) visually ensures the correctness of the inverse relationship.