Problem 44 - 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 \(f\left(f^{-1}(x)\right)=x\) $$ f(x)=\left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} $$

Step-by-Step Solution

Answer

The inverse formula is \( f^{-1}(x) = \sqrt[3]{\frac{2-\sqrt[5]{x}}{\sqrt[5]{x}-1}} \). Both inverse property checks are satisfied.

Step 1: Determine the Range of the Original Function

The function given is \( f(x)=\left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} \). Before finding the inverse, examine the range of its simpler form \( g(x) = \frac{x^3 + 2}{x^3 + 1} \). If \( y = g(x) \), we can rearrange \( g(x) \) as \( y \left( x^3+1 \right) = x^3 + 2 \), leading to \( x^3(y - 1) = 2 - y \). From this expression, understand that \( g(x) \) is always greater than or equal to 1, meaning \( f(x) \) starts from 1 onward.

Step 2: Substitute and Simplify for Inverse Function

To find the inverse, start by setting \( y = f(x) \). This gives \[ y = \left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} \]. Solving for \( x \), we take the fifth root, then rearrange to obtain: \[ \sqrt[5]{y} = \frac{x^3 + 2}{x^3 + 1} \]. Cross-multiply to get \( \sqrt[5]{y}(x^3 + 1) = x^3 + 2 \). Expand and rearrange to find \[ x^3(\sqrt[5]{y} - 1) = 2 - \sqrt[5]{y} \]. Finally, \( x^3 = \frac{2 - \sqrt[5]{y}}{\sqrt[5]{y} - 1} \) which simplifies to \( x = \sqrt[3]{\frac{2 - \sqrt[5]{y}}{\sqrt[5]{y} - 1}} \).

Step 3: Simplify to Find \( f^{-1}(x) \)

Considering \( y = f(x) \) and now solving for \( x \), this implies \( x \) itself becomes \( f^{-1}(y) \), so we replace \( y \) with \( x \). Thus, \[ f^{-1}(x) = \sqrt[3]{\frac{2 - \sqrt[5]{x}}{\sqrt[5]{x} - 1}} \].

Step 4: Verify \( f^{-1}(f(x)) = x \)

Substitute \( f(x) \) into \( f^{-1} \) and simplify: \[ f^{-1}(f(x)) = \sqrt[3]{\frac{2 - \sqrt[5]{\left( \left(\frac{x^3 + 2}{x^3 + 1}\right)^5 \right)}}{\sqrt[5]{\left( \left(\frac{x^3 + 2}{x^3 + 1}\right)^5 \right)} - 1}} \]. Simplifying further, the expression inside the cube root collapses to \( x^3 \), so \( f^{-1}(f(x)) = x \).

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

Substitute \( f^{-1}(x) \) into \( f \): \[ f(f^{-1}(x)) = \left( \frac{(\sqrt[3]{\frac{2-\sqrt[5]{x}}{\sqrt[5]{x}-1}})^3 + 2}{(\sqrt[3]{\frac{2-\sqrt[5]{x}}{\sqrt[5]{x}-1}})^3 + 1} \right)^5 \]. The inner expression simplifies to \( x \), resulting in \( f(f^{-1}(x)) = x \).

Key Concepts

Algebraic Manipulation

To find the inverse of a function, you need solid algebraic manipulation skills. When dealing with complex functions like \( f(x)=\left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} \), the task involves various algebraic steps to isolate \( x \) in terms of \( y \).

For this function, the first step is to set \( y = f(x) \). This gives us \( y = \left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} \). To solve for \( x \), we take the fifth root of both sides: \( \sqrt[5]{y} = \frac{x^3 + 2}{x^3 + 1} \).

This expression can then be manipulated through cross-multiplication, yielding \( \sqrt[5]{y}(x^3 + 1) = x^3 + 2 \). Further algebraic steps lead to solving for \( x^3 \), and finally \( x \).

These manipulations result in the inverse function: \( f^{-1}(x) = \sqrt[3]{\frac{2 - \sqrt[5]{x}}{\sqrt[5]{x} - 1}} \).

Understanding that each step involves applying reverse operations like taking roots, isolating variables, and reorganizing equations is key to mastering function inversions.

Range of a Function

Understanding the range of a function is crucial for many mathematical tasks, including finding inverses. Before diving into inverses, we must determine what values the function can output.

For the function \( f(x)=\left(\frac{x^{3}+2}{x^{3}+1}\right)^{5} \), the first step is to find the range of the simpler form \( g(x)=\frac{x^{3}+2}{x^{3}+1}\). If we set \( y = g(x) \), rearranging gives: \( x^3(y - 1) = 2 - y \), indicating that \( g(x) \) is always greater than or equal to 1.

This means the range of \( f(x) \), which is \( g(x)^5 \), will start from 1 onward without bound. This is because any value \( g(x) \) greater or equal to 1 will lead to \( f(x) \) also being greater or equal to 1.

Knowing the range helps to frame the domain of the inverse function, ensuring correct interpretation and solving of the inverse.

Function Verification

To verify if functions are indeed true inverses, we need to check both \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(x)) = x \). This dual verification ensures that the inverse found works correctly in both directions.

Starting with \( f^{-1}(f(x)) \), substituting \( f(x) \) into \( f^{-1} \) involves plugging in the value of \( f(x) = \left(\frac{x^3 + 2}{x^3 + 1}\right)^5 \) into the inverse equation. This process reduces the expression down to \( x \), confirming the first verification.

The next step is to ensure that \( f(f^{-1}(x)) = x \). You substitute \( f^{-1}(x) \) back into the original function \( f \). The algebraic simplification of this combination also resolves to \( x \), which confirms our function's accuracy.

Therefore, completing these verification steps guarantees the accuracy of our inverse function and provides confidence that our algebraic manipulations and range considerations were correct from the start.