Question

Find f superscript negative 1 baseline for the function f left parenthesis x right parenthesis equals root index 3 start root 2 x minus 1 end root.
(1 point)
Responses

f superscript negative 1 baseline left parenthesis x right parenthesis equals one half left parenthesis x cubed plus 1 right parenthesis
Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals one half left parenthesis x cubed plus 1 right parenthesis

f superscript negative 1 baseline left parenthesis x right parenthesis equals one half left parenthesis x plus 1 right parenthesis cubed
Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals one half left parenthesis x plus 1 right parenthesis cubed

f superscript negative 1 baseline left parenthesis x right parenthesis equals left parenthesis start fraction x over 2 end fraction plus 1 right parenthesis cubed
Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals left parenthesis start fraction x over 2 end fraction plus 1 right parenthesis cubed

f superscript negative 1 baseline left parenthesis x right parenthesis equals left parenthesis start fraction x over 2 end fraction right parenthesis cubed plus 1
Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals left parenthesis start fraction x over 2 end fraction right parenthesis cubed plus 1

Answer 1

To find the inverse function \( f^{-1}(x) \) for the given function \( f(x) = \sqrt[3]{2x - 1} \), follow these steps:

1. Start by writing the function:
\[
y = \sqrt[3]{2x - 1}
\]

2. Solve for \( x \) in terms of \( y \):
- First, cube both sides to eliminate the cube root:
\[
y^3 = 2x - 1
\]
- Next, solve for \( 2x \):
\[
2x = y^3 + 1
\]
- Finally, solve for \( x \):
\[
x = \frac{y^3 + 1}{2}
\]

3. Now, replace \( y \) with \( x \) to express the inverse:
\[
f^{-1}(x) = \frac{x^3 + 1}{2}
\]

Now, looking at the provided options:

1. \( f^{-1}(x) = \frac{1}{2}(x^3 + 1) \) - This is equivalent to our derived equation.
2. \( f^{-1}(x) = \frac{1}{2}(x + 1)^3 \) - This suggests a different expression.
3. \( f^{-1}(x) = \left(\frac{x}{2} + 1\right)^3 \) - Also suggests a different expression.
4. \( f^{-1}(x) = \left(\frac{x}{2}\right)^3 + 1 \) - Not consistent with our derived function.

The correct answer is:
\[
f^{-1}(x) = \frac{1}{2}(x^3 + 1)
\]