To evaluate the function [tex]\( h(x) \)[/tex] at the given points, let's analyze the piecewise defined function and the appropriate intervals for each given value.
The function is defined as follows:
[tex]\[
h(x)=\left\{\begin{array}{ll}
-x^2-6 x-9, & x<-2 \\
\left(\frac{1}{3}\right)^x-4, & -2 \leq x \leq 2 \\
\frac{1}{2} x-4, & x>2
\end{array}\right.
\][/tex]
1. Evaluating [tex]\( h(-3) \)[/tex]:
- Since [tex]\(-3 < -2\)[/tex], we use the first piece of the function:
[tex]\[
h(x) = -x^2 - 6x - 9
\][/tex]
Substituting [tex]\( x = -3 \)[/tex]:
[tex]\[
h(-3) = -(-3)^2 - 6(-3) - 9
\][/tex]
Simplifying:
[tex]\[
= -9 + 18 - 9 = 0
\][/tex]
Therefore,
[tex]\[
h(-3) = 0
\][/tex]
2. Evaluating [tex]\( h(-2) \)[/tex]:
- Since [tex]\(-2 \leq -2 \leq 2\)[/tex], we use the second piece of the function:
[tex]\[
h(x) = \left(\frac{1}{3}\right)^x - 4
\][/tex]
Substituting [tex]\( x = -2 \)[/tex]:
[tex]\[
h(-2) = \left(\frac{1}{3}\right)^{-2} - 4
\][/tex]
Simplifying:
[tex]\[
= 3^2 - 4 = 9 - 4 = 5
\][/tex]
Therefore,
[tex]\[
h(-2) = 5
\][/tex]
3. Evaluating [tex]\( h(4) \)[/tex]:
- Since [tex]\(4 > 2\)[/tex], we use the third piece of the function:
[tex]\[
h(x) = \frac{1}{2}x - 4
\][/tex]
Substituting [tex]\( x = 4 \)[/tex]:
[tex]\[
h(4) = \frac{1}{2}(4) - 4
\][/tex]
Simplifying:
[tex]\[
= 2 - 4 = -2
\][/tex]
Therefore,
[tex]\[
h(4) = -2
\][/tex]
Summarizing the results:
[tex]\(
\begin{array}{l}
h(-3)= 0 \\
h(-2)= 5 \\
h(4)= -2
\end{array}
\)[/tex]
