Question

question
for the function $f(x) = \frac{(x - 9)^{\frac{1}{3}}}{5}$, find $f^{-1}(x)$.

answer
$\circ$ $f^{-1}(x) = (5x + 9)^3$ $\circ$ $f^{-1}(x) = (5(x + 9))^3$
$\circ$ $f^{-1}(x) = 5(x + 9)^3$ $\circ$ $f^{-1}(x) = (5x)^3 + 9$

Explanation:

Step1: Let \( y = f(x) \)

Set \( y=\frac{(x - 9)^{\frac{1}{3}}}{5} \).

Step2: Solve for \( x \) in terms of \( y \)

First, multiply both sides by \( 5 \): \( 5y=(x - 9)^{\frac{1}{3}} \).
Then, cube both sides to eliminate the cube root: \( (5y)^{3}=x - 9 \).
Finally, add \( 9 \) to both sides: \( x=(5y)^{3}+9 \)? Wait, no, wait. Wait, \( (5y)^{\frac{1}{3}\times3}=x - 9 \)? Wait, no, original step: \( 5y=(x - 9)^{\frac{1}{3}} \), cubing both sides: \( (5y)^{3}=x - 9 \)? No, wait, \( (a)^{\frac{1}{3}}=b \) implies \( a = b^{3} \). So \( x - 9=(5y)^{3} \)? No, wait, \( 5y=(x - 9)^{\frac{1}{3}} \), so cubing both sides: \( (5y)^{3}=x - 9 \)? No, that's wrong. Wait, \( (x - 9)^{\frac{1}{3}}=5y \), so raise both sides to the power of \( 3 \): \( x - 9=(5y)^{3} \)? No, \( (a^{\frac{1}{3}})^{3}=a \), so \( x - 9=(5y)^{3} \)? Wait, no, \( (5y)^{3}=125y^{3} \), but that's not right. Wait, no, let's do it again. Let \( y=\frac{(x - 9)^{\frac{1}{3}}}{5} \). Multiply both sides by \( 5 \): \( 5y=(x - 9)^{\frac{1}{3}} \). Now, cube both sides: \( (5y)^{3}=x - 9 \)? No, \( (5y)^{3}=125y^{3} \), but actually, \( (a)^{\frac{1}{3}}=b \) so \( a = b^{3} \). So \( x - 9=(5y)^{3} \)? Wait, no, \( 5y=(x - 9)^{\frac{1}{3}} \), so \( x - 9=(5y)^{3} \)? Then \( x=(5y)^{3}+9 \)? But that's not one of the options. Wait, no, wait, I made a mistake. Wait, the original function is \( f(x)=\frac{(x - 9)^{\frac{1}{3}}}{5} \), so let's write \( y = \frac{(x - 9)^{\frac{1}{3}}}{5} \). Multiply both sides by \( 5 \): \( 5y=(x - 9)^{\frac{1}{3}} \). Now, cube both sides: \( (5y)^{3}=x - 9 \)? No, \( (5y)^{\frac{1}{3}\times3}=x - 9 \)? Wait, no, \( (a^{\frac{1}{3}})^{3}=a \), so \( x - 9=(5y)^{3} \)? Then \( x=(5y)^{3}+9 \)? But the options are:

1. \( f^{-1}(x)=(5x + 9)^{3} \)
2. \( f^{-1}(x)=(5(x + 9))^{3} \)
3. \( f^{-1}(x)=5(x + 9)^{3} \)
4. \( f^{-1}(x)=(5x)^{3}+9 \)

Wait, wait, maybe I messed up the steps. Let's start over. Let \( y = f(x)=\frac{(x - 9)^{\frac{1}{3}}}{5} \).

Step 1: Swap \( x \) and \( y \): \( x=\frac{(y - 9)^{\frac{1}{3}}}{5} \).

Step 2: Solve for \( y \). Multiply both sides by \( 5 \): \( 5x=(y - 9)^{\frac{1}{3}} \).

Step 3: Cube both sides: \( (5x)^{3}=y - 9 \).

Step 4: Add \( 9 \) to both sides: \( y=(5x)^{3}+9 \). Wait, but that's option 4? But let's check the options again. Wait, no, wait, maybe I made a mistake in the cube. Wait, \( (5x)^{\frac{1}{3}\times3}=y - 9 \)? No, \( (a^{\frac{1}{3}})^{3}=a \), so \( y - 9=(5x)^{3} \), so \( y=(5x)^{3}+9 \), which is option 4: \( f^{-1}(x)=(5x)^{3}+9 \). Wait, but let's check with the options. Wait, no, maybe I messed up the original function. Wait, the original function is \( f(x)=\frac{(x - 9)^{\frac{1}{3}}}{5} \), so inverse:

Let \( y = \frac{(x - 9)^{\frac{1}{3}}}{5} \)

Swap \( x \) and \( y \): \( x = \frac{(y - 9)^{\frac{1}{3}}}{5} \)

Multiply both sides by \( 5 \): \( 5x = (y - 9)^{\frac{1}{3}} \)

Cube both sides: \( (5x)^3 = y - 9 \)

Add \( 9 \): \( y = (5x)^3 + 9 \)

So \( f^{-1}(x)=(5x)^3 + 9 \), which is the fourth option. Wait, but let's check the options:

Option 4: \( f^{-1}(x)=(5x)^{3}+9 \)

Yes, that's correct. Wait, but let's verify with an example. Let's take \( x = 9 + 1 = 10 \), then \( f(10)=\frac{(10 - 9)^{\frac{1}{3}}}{5}=\frac{1^{\frac{1}{3}}}{5}=\frac{1}{5} \). Now, apply \( f^{-1}(\frac{1}{5}) \): \( (5\times\frac{1}{5})^3 + 9=(1)^3 + 9 = 10 \), which is correct. So the correct answer is \( f^{-1}(x)=(5x)^{3}+9 \).

Answer:

\( f^{-1}(x)=(5x)^{3}+9 \) (the fourth option)