Question

suppose $f(x) = \frac{x^3 - 7}{10}$ and $g(x) = \sqrt{x + 1}$. calculate $(g \circ f^{-1})(2)$ \
\bigcirc 4 \
\bigcirc 3 \
\bigcirc 2 \
\bigcirc 1 \
\bigcirc 0

Explanation:

Step1: Recall the composition of inverse functions

We know that \((g\circ f^{-1})(2)=g(f^{-1}(2))\). By the definition of an inverse function, if \(y = f(x)\), then \(f^{-1}(y)=x\) when \(f(x)=y\). So we first need to find \(x\) such that \(f(x) = 2\), because \(f^{-1}(2)=x\) implies \(f(x)=2\).

Step2: Solve \(f(x)=2\) for \(x\)

Given \(f(x)=\frac{x^{3}-7}{10}\), set \(\frac{x^{3}-7}{10}=2\).
Multiply both sides by 10: \(x^{3}-7 = 2\times10=20\).
Add 7 to both sides: \(x^{3}=20 + 7=27\).
Take the cube root of both sides: \(x=\sqrt[3]{27}=3\). So \(f^{-1}(2)=3\) (since \(f(3)=2\)).

Step3: Evaluate \(g(f^{-1}(2))\)

Now we need to find \(g(3)\) because \(f^{-1}(2)=3\). Given \(g(x)=\sqrt{x + 1}\), substitute \(x = 3\) into \(g(x)\):
\(g(3)=\sqrt{3+1}=\sqrt{4}=2\).

Answer:

2