Question

1. let $f(x) = \sqrt{x + 1}$, $g(x) = 2x - 5$, and $h(x) = 3x^2 - 3$. find the indicated value. $g(h(-2)) = \square$

Explanation:

Step1: Calculate \( h(-2) \)

First, we need to find the value of \( h(x) \) when \( x = -2 \). The function \( h(x) = 3x^2 - 3 \). Substitute \( x = -2 \) into \( h(x) \):
\[
h(-2)=3(-2)^2 - 3
\]
Calculate \( (-2)^2 = 4 \), then:
\[
h(-2)=3\times4 - 3=12 - 3 = 9
\]

Step2: Calculate \( g(h(-2)) \)

Now that we know \( h(-2)=9 \), we need to find \( g(9) \) since \( g(h(-2)) = g(9) \). The function \( g(x)=2x - 5 \). Substitute \( x = 9 \) into \( g(x) \):
\[
g(9)=2\times9 - 5
\]
Calculate \( 2\times9 = 18 \), then:
\[
g(9)=18 - 5 = 13
\]

Answer:

\( 13 \)