For the function f, of, x, equals, start fraction, x, start superscript, one third, end superscript, divided by, 5, end fraction, plus, 7f(x)=

Question

5
x
3
1
​


​
+7, find f, to the power minus 1 , left parenthesis, x, right parenthesisf
−1
(x).
Answer
Attempt 1 out of 2
Multiple Choice Answers
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, left parenthesis, 5, x, minus, 7, right parenthesis, cubedf
−1
(x)=(5x−7)
3

f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, left parenthesis, 5, x, right parenthesis, cubed, minus, 7f
−1
(x)=(5x)
3
−7
f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, left parenthesis, 5, left parenthesis, x, minus, 7, right parenthesis, right parenthesis, cubedf
−1
(x)=(5(x−7))
3

f, to the power minus 1 , left parenthesis, x, right parenthesis, equals, 5, left parenthesis, x, minus, 7, right parenthesis, cubedf
−1
(x)=5(x−7)
3

Answer 1

To find the inverse function $f^{-1}(x)$ for the given function

$$f(x)=\frac{x^{1/3}}{5}+7,$$

we want to express $x$ in terms of $y$ where $y=f(x)$.

Let's start by setting $y=f(x)$:

$$y=\frac{x^{1/3}}{5}+7$$

Now, we solve for $x$:

1. Subtract 7 from both sides:

$$y-7=\frac{x^{1/3}}{5}$$

2. Multiply both sides by 5:

$$5(y-7)=x^{1/3}$$

3. Now, cube both sides to eliminate the cube root:

$$x=[5(y-7){]}^3$$

This means we can express $f^{-1}(x)$ as:

$$f^{-1}(x)=[5(x-7){]}^3$$

Thus, the correct multiple choice answer is:

$$f^{-1}(x)=(5(x-7){)}^3$$