The function $$ f(x)=x^{3}+7 $$ is one-to-one. a. Find an equation for $$ f^{-1} $$, the inverse function. b. Verify that your equation is correct by showing that $$ f\left(f^{-1}(x)\right)=x $$ and $$ f^{-1}(f(x))=x $$.

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### Problem

The function $$ f(x) = x^{3} + 7 $$ is one-to-one.

**a.** Find an equation for $$ f^{-1}(x) $$, the inverse function.

**b.** Verify that your equation is correct by showing that $$ f\left(f^{-1}(x)\right) = x $$ and $$ f^{-1}\left(f(x)\right) = x $$.

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### Solution

#### Part a: Finding the Inverse Function $$ f^{-1}(x) $$

To find the inverse function $$ f^{-1}(x) $$ of the given function $$ f(x) = x^{3} + 7 $$, follow these steps:

1. **Replace $$ f(x) $$ with $$ y $$:**
   $$
   y = x^{3} + 7
   $$

2. **Swap $$ x $$ and $$ y $$ to prepare for solving for $$ y $$:**
   $$
   x = y^{3} + 7
   $$

3. **Solve for $$ y $$:**
   $$
   x - 7 = y^{3}
   $$
   $$
   y = \sqrt[3]{x - 7}
   $$

4. **Express the inverse function:**
   $$
   f^{-1}(x) = \sqrt[3]{x - 7}
   $$
   Alternatively, it can be written using fractional exponents:
   $$
   f^{-1}(x) = (x - 7)^{1/3}
   $$

#### Part b: Verifying the Inverse Function

To verify that $$ f^{-1}(x) $$ is indeed the inverse of $$ f(x) $$, we need to show that:
1. $$ f\left(f^{-1}(x)\right) = x $$
2. $$ f^{-1}\left(f(x)\right) = x $$

**1. Verifying $$ f\left(f^{-1}(x)\right) = x $$:**

Start with $$ f^{-1}(x) = \sqrt[3]{x - 7} $$:
$$
f\left(f^{-1}(x)\right) = f\left(\sqrt[3]{x - 7}\right) = \left(\sqrt[3]{x - 7}\right)^{3} + 7
$$
Simplify the expression:
$$
\left(\sqrt[3]{x - 7}\right)^{3} = x - 7
$$
$$
f\left(f^{-1}(x)\right) = (x - 7) + 7 = x
$$
Thus, $$ f\left(f^{-1}(x)\right) = x $$.

**2. Verifying $$ f^{-1}\left(f(x)\right) = x $$:**

Start with $$ f(x) = x^{3} + 7 $$:
$$
f^{-1}\left(f(x)\right) = f^{-1}\left(x^{3} + 7\right) = \sqrt[3]{(x^{3} + 7) - 7} = \sqrt[3]{x^{3}} = x
$$
Thus, $$ f^{-1}\left(f(x)\right) = x $$.

Both compositions yield $$ x $$, confirming that $$ f^{-1}(x) = \sqrt[3]{x - 7} $$ is indeed the inverse function of $$ f(x) = x^{3} + 7 $$.

### Answer

**a.** The inverse function is $$ f^{-1}(x) = \sqrt[3]{\,x - 7\,} $$.

**b.** Verification:

$$
f\left(f^{-1}(x)\right) = f\left(\sqrt[3]{\,x - 7\,}\right) = \left(\sqrt[3]{\,x - 7\,}\right)^3 + 7 = x - 7 + 7 = x
$$

$$
f^{-1}\left(f(x)\right) = f^{-1}\left(x^{3} + 7\right) = \sqrt[3]{\,x^{3} + 7 - 7\,} = \sqrt[3]{\,x^{3}\,} = x
$$

Both compositions simplify to $$ x $$, confirming that $$ f^{-1}(x) = \sqrt[3]{\,x - 7\,} $$ is the correct inverse of $$ f(x) = x^{3} + 7 $$.
**a.** The inverse function is $$ f^{-1}(x) = \sqrt[3]{x - 7} $$. **b.** Verification: $$ f\left(f^{-1}(x)\right) = x $$ $$ f^{-1}\left(f(x)\right) = x $$ Both compositions equal $$ x $$, confirming that $$ f^{-1}(x) = \sqrt[3]{x - 7} $$ is the correct inverse of $$ f(x) = x^{3} + 7 $$.

The function \( f(x)=x^{3}+7 \) is one-to-one. a. Find an equation for \( f^{-1} \), the inverse function. b. Verify that your equation is correct by showing that \( f\left(f^{-1}(x)\right)=x \) and \( f^{-1}(f(x))=x \).

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