Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a formula for \(f^{-1}(x)\) and then verify that \(f^{-1}(f(x))=x\) and \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=x^{5 / 2}, x \geq 0 $$

Step-by-Step Solution

Answer

The inverse function is \(f^{-1}(x) = x^{2/5}\). It satisfies both properties.

Step 1: Understand the problem

We need to find the inverse function \(f^{-1}(x)\) for the given function \(f(x) = x^{5/2}\) where \(x \geq 0\). Then, we will verify the properties \(f^{-1}(f(x)) = x\) and \(f(f^{-1}(x)) = x\).

Step 2: Express the function with \(y\)

Start by expressing the function \(f(x)\) using a different variable: \(y = x^{5/2}\). Our goal is to solve for \(x\) in terms of \(y\) to find the inverse.

Step 3: Solve for \(x\)

Isolate \(x\) in the equation \(y = x^{5/2}\). Raise both sides to the power of \(2/5\) to get: \(x = y^{2/5}\).

Step 4: Define the inverse function

Based on the rearranged equation, define the inverse function as: \(f^{-1}(x) = x^{2/5}\).

Step 5: Verify \(f^{-1}(f(x)) = x\)

Substitute \(f(x)\) into \(f^{-1}(x)\):\[f^{-1}(f(x)) = f^{-1}(x^{5/2}) = (x^{5/2})^{2/5}\]This simplifies to:\[x^{(5/2) \cdot (2/5)} = x^1 = x\]Thus, \(f^{-1}(f(x)) = x\) is verified.

Step 6: Verify \(f(f^{-1}(x)) = x\)

Substitute \(f^{-1}(x)\) into \(f(x)\):\[f(f^{-1}(x)) = f(x^{2/5}) = (x^{2/5})^{5/2}\]This simplifies to:\[x^{(2/5) \cdot (5/2)} = x^1 = x\]Thus, \(f(f^{-1}(x)) = x\) is verified.

Key Concepts

Function Composition

Function composition is like applying one function to the output of another function. This means if you have two functions, say, \(f(x)\) and \(g(x)\), the composition \(f(g(x))\) involves plugging \(g(x)\) into \(f(x)\). It’s written as \((f \circ g)(x) = f(g(x))\).

The idea behind composition is to create complex operations by combining simpler ones. For example, if you have \(g(x) = x^2\) and \(f(x) = x + 3\), composing them, \(f(g(x))\) means substituting \(g(x)\) in \(f(x)\). This becomes \((x^2) + 3 = x^2 + 3\).

In the context of inverse functions, we often check if two functions are inverses by composing them and ensuring that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). If this holds true, the functions are confirmed to be inverses of each other.

Verification of Inverse Functions

To verify if two functions are truly inverses, we need to check if composing them yields the identity function. We use the properties:

• \(f(f^{-1}(x)) = x\)
• \(f^{-1}(f(x)) = x\)

These properties reveal that applying one function and then its inverse returns you to the starting value, \(x\).

In our exercise, we verified this by substituting \(f(x)\) into the proposed inverse function \(f^{-1}(x)\) and simplifying. For the given function \(f(x) = x^{5/2}\) and its proposed inverse \(f^{-1}(x) = x^{2/5}\), we showed that \(f(f^{-1}(x)) = x\) by calculating \((x^{2/5})^{5/2} = x\). Likewise, we proved \(f^{-1}(f(x)) = x\) with \((x^{5/2})^{2/5} = x\).

Both computations resulted in \(x\), thereby confirming that \(f(x)\) and \(f^{-1}(x)\) are indeed inverses.

Power Functions

Power functions are a special type of function where the variable \(x\) is raised to a constant power, represented as \(f(x) = x^a\). These functions are fundamental in algebra and calculus due to their simple yet versatile form.

In the given exercise, we dealt with a power function \(f(x) = x^{5/2}\). Here, the base is \(x\) and the exponent is a rational number, \(\frac{5}{2}\). This type of power function is continuous and differentiable where \(x\geq0\), making it straightforward to find inverses.

To find the inverse, we need to "undo" the power by altering the exponent. This means using the reciprocal of \(\frac{5}{2}\), which is \(\frac{2}{5}\). By raising both sides of \(y = x^{5/2}\) to \(\frac{2}{5}\), we isolated \(x\) and found that \(f^{-1}(x) = x^{2/5}\).

Power functions are not only essential for inverses but provide a basis for exploring growth patterns and transformations in mathematics.