Question

use the selected values of a linear function g(x) in the table and the equation h(x) shown below to evaluate (g ∘ h)(-3).

x	g(x)
-3	-5
-1	-1
0	1
4	9

$h(x) = x^2 - 2$
\\(\circ\\) -22
\\(\circ\\) -21
\\(\circ\\) 14
\\(\circ\\) 15

Explanation:

Step1: Recall the composition of functions

The composition \((g \circ h)(-3)\) means \(g(h(-3))\). So first, we need to find \(h(-3)\).

Step2: Calculate \(h(-3)\)

Given \(h(x)=x^{2}-2\), substitute \(x = - 3\) into \(h(x)\):
\(h(-3)=(-3)^{2}-2\)
\(=9 - 2\)
\(=7\)

Step3: Calculate \(g(h(-3))=g(7)\)? Wait, no, wait. Wait, the table for \(g(x)\): we need to check if we made a mistake. Wait, no, \(h(-3) = 7\)? Wait, no, wait, the table of \(g(x)\) has \(x=-3,-1,0,4\). Wait, maybe I miscalculated \(h(-3)\)? Wait, no, \(h(x)=x^{2}-2\), so \(h(-3)=(-3)^2 - 2=9 - 2 = 7\). But the table of \(g(x)\) does not have \(x = 7\). Wait, that can't be. Wait, maybe I misread the problem. Wait, the problem is \((g\circ h)(-3)=g(h(-3))\). Wait, maybe I made a mistake in \(h(-3)\). Wait, no, let's re - check. \(h(x)=x^{2}-2\), \(x=-3\): \((-3)^2=9\), \(9 - 2 = 7\). But the table of \(g(x)\) has \(x\) values \(-3,-1,0,4\). Wait, maybe the problem is \((h\circ g)(-3)\)? No, the problem says \((g\circ h)(-3)\). Wait, maybe I made a mistake. Wait, no, let's check the table again. Wait, the table is for \(g(x)\): when \(x=-3\), \(g(x)=-5\); \(x = - 1\), \(g(x)=-1\); \(x = 0\), \(g(x)=1\); \(x = 4\), \(g(x)=9\). Wait, maybe I messed up the order of composition. Wait, \((g\circ h)(-3)=g(h(-3))\). So \(h(-3)=7\), but \(g(7)\) is not in the table. Wait, that must mean I made a mistake. Wait, no, wait, maybe the function \(g(x)\) is linear. Let's find the equation of \(g(x)\) since it's a linear function. Let's find the slope first. Take two points, say \((-3,-5)\) and \((-1,-1)\). The slope \(m=\frac{-1-(-5)}{-1 - (-3)}=\frac{4}{2}=2\). So the equation of \(g(x)\) is \(y - (-5)=2(x - (-3))\), so \(y + 5=2(x + 3)\), \(y=2x+6 - 5\), \(y = 2x + 1\). Let's check with \(x = 0\): \(y=2(0)+1 = 1\), which matches the table. \(x = 4\): \(y=2(4)+1=9\), which also matches. So \(g(x)=2x + 1\). Now, \(h(-3)=(-3)^2-2 = 7\), then \(g(7)=2(7)+1=15\)? Wait, but let's check again. Wait, no, the problem is \((g\circ h)(-3)\). So first, \(h(-3)=(-3)^2-2=7\), then \(g(7)\). Since \(g(x)\) is linear with \(g(x)=2x + 1\), \(g(7)=2\times7+1 = 15\). Wait, but let's check the options. 15 is an option.

Wait, maybe I should have found the equation of \(g(x)\) first. Let's redo the steps correctly:

Step1: Find the equation of \(g(x)\)

Since \(g(x)\) is linear, let \(g(x)=mx + b\).
Using the points \((-3,-5)\) and \((-1,-1)\):
For \(x=-3,y = - 5\): \(-5=-3m + b\)
For \(x=-1,y=-1\): \(-1=-m + b\)
Subtract the first equation from the second equation:
\((-m + b)-(-3m + b)=-1-(-5)\)
\(-m + b + 3m - b=4\)
\(2m=4\)
\(m = 2\)
Substitute \(m = 2\) into \(-1=-m + b\):
\(-1=-2 + b\)
\(b=1\)
So \(g(x)=2x + 1\)

Step2: Calculate \(h(-3)\)

\(h(x)=x^{2}-2\), so \(h(-3)=(-3)^{2}-2=9 - 2 = 7\)

Step3: Calculate \(g(h(-3))=g(7)\)

Substitute \(x = 7\) into \(g(x)=2x + 1\):
\(g(7)=2\times7+1=14 + 1=15\)

Answer:

15