For the function f, of, x, equals, 9, x, to the power 5f(x)=9x

Question

5
, find f, to the power minus 1 , left parenthesis, x, right parenthesisf
−1
(x).
Answer
Multiple Choice Answers
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, start root, start index, 5, end index, left parenthesis, start fraction, x, divided by, 9, end fraction, right parenthesis, end rootf
−1
(x)=
5

(
9
x
​
)
​

f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, 9, start root, start index, 5, end index, x, end rootf
−1
(x)=9
5

x
​

f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, start fraction, start root, start index, 5, end index, x, end root, divided by, 9, end fractionf
−1
(x)=
9
5

x
​

​

f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, start fraction, x, to the power 5 , divided by, 9, end fractionf
−1
(x)=
9
x
5

​

Answer 1

To find the inverse of the function \( f(x) = 9x^5 \), we follow these steps:

1. Start by replacing \( f(x) \) with \( y \): \[ y = 9x^5 \]

2. Next, solve for \( x \) in terms of \( y \): \[ y = 9x^5 \quad \Rightarrow \quad \frac{y}{9} = x^5 \]

3. Take the fifth root of both sides: \[ x = \left( \frac{y}{9} \right)^{\frac{1}{5}} \]

4. Now replace \( x \) with \( f^{-1}(y) \): \[ f^{-1}(y) = \left( \frac{y}{9} \right)^{\frac{1}{5}} \]

5. Finally, replace \( y \) with \( x \) to express it in the standard form: \[ f^{-1}(x) = \left( \frac{x}{9} \right)^{\frac{1}{5}} \]

Therefore, converting it to a root notation: \[ f^{-1}(x) = \sqrt[5]{\frac{x}{9}} \]

Now, checking your multiple choice answers, the correct choice corresponds to: \[ f^{-1}(x) = \sqrt[5]{\frac{x}{9}} \]