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Question
question 1
1 pts
which function is the inverse of:
$g(x) = \frac{(x - 3)^3}{2} - 2$ ?
$\bigcirc$ $f(x) = 2\sqrt3{x + 1} + 3$
$\bigcirc$ $f(x) = \sqrt3{2x + 6} - 2$
$\bigcirc$ $f(x) = 2\sqrt3{x + 3} + 1$
$\bigcirc$ $f(x) = \sqrt3{2x + 4} + 3$
question 2
1 pts
Step1: Replace \( g(x) \) with \( y \)
We start with the function \( g(x)=\frac{(x - 3)^{3}}{2}-2 \). Replace \( g(x) \) with \( y \), so we have \( y=\frac{(x - 3)^{3}}{2}-2 \).
Step2: Swap \( x \) and \( y \)
To find the inverse, we swap \( x \) and \( y \). This gives us \( x=\frac{(y - 3)^{3}}{2}-2 \).
Step3: Solve for \( y \)
First, add 2 to both sides of the equation:
\( x + 2=\frac{(y - 3)^{3}}{2} \)
Then, multiply both sides by 2:
\( 2(x + 2)=(y - 3)^{3} \)
Simplify the left side: \( 2x+4=(y - 3)^{3} \)
Take the cube root of both sides:
\( \sqrt[3]{2x + 4}=y - 3 \)
Finally, add 3 to both sides to solve for \( y \):
\( y=\sqrt[3]{2x + 4}+3 \)
So the inverse function \( f(x)=\sqrt[3]{2x + 4}+3 \)
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\( f(x)=\sqrt[3]{2x + 4}+3 \) (the fourth option: \( \boldsymbol{f(x)=\sqrt[3]{2x + 4}+3} \))