Question

given the function $f(x)=\frac{1}{2}\sqrt{x}$, use $f(g(x))$ to verify or disprove the proposed inverse function $g(x)=4x^2$.
the function g(x) select the inverse of f(x) because f(g(x)) select

Explanation:

Step1: Substitute $g(x)$ into $f(x)$

$f(g(x)) = \frac{1}{2}\sqrt{4x^2}$

Step2: Simplify the square root term

$\sqrt{4x^2} = 2|x|$, so $f(g(x)) = \frac{1}{2} \times 2|x| = |x|$

Step3: Compare to identity function

For $g(x)$ to be the inverse of $f(x)$, $f(g(x))$ must equal $x$ (for the domain where $f(x)$ is defined, $x\geq0$, $|x|=x$ but we also need to check the full condition, and here $f(g(x))=|x|$, which is not equal to $x$ for negative $x$, and even for non-negative $x$, we also need to check $g(f(x))$, but from $f(g(x))$ alone, it does not satisfy the inverse requirement fully as it gives absolute value, not just $x$ for all valid inputs, and the standard inverse check requires $f(g(x))=x$ and $g(f(x))=x$ for all $x$ in the respective domains. For $f(x)=\frac{1}{2}\sqrt{x}$, domain is $x\geq0$, range is $y\geq0$. $g(x)=4x^2$, if we compute $g(f(x))=4(\frac{1}{2}\sqrt{x})^2=4\times\frac{1}{4}x=x$, but $f(g(x))=|x|$, which is $x$ only when $x\geq0$, but for the inverse, we need $f(g(x))=x$ for all $x$ in the domain of $g(x)$ that maps to the domain of $f(x)$. However, since $f(g(x))$ is not equal to $x$ for negative $x$, and the strict inverse condition is not met fully, $g(x)$ is not the inverse.

Answer:

The function g(x) is not the inverse of f(x) because $f(g(x)) = |x|
eq x$ (for all real $x$, and does not satisfy the requirement that $f(g(x))=x$ for all relevant inputs).