To evaluate the piecewise function [tex]\( h(x) \)[/tex], we look at each interval separately, using the described function forms. Let's determine [tex]\( h \)[/tex] at specific values of [tex]\( x \)[/tex] in each relevant interval.
Given function:
[tex]\[
h(x)=\left\{
\begin{array}{ll}
-x^2-6x-9, & x<-2 \\
\left(\frac{1}{3}\right)^x-4, & -2 \leq x \leq 2 \\
\frac{1}{2} x-4, & x>2
\end{array}
\right.
\][/tex]
1. Calculate [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]\[
h(-3) = -(-3)^2 - 6(-3) - 9 = -9 + 18 - 9 = 0
\][/tex]
So, [tex]\( h(-3) = 0 \)[/tex].
2. Calculate [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]\[
h(-2) = \left(\frac{1}{3}\right)^{-2} - 4 = 3^2 - 4 = 9 - 4 = 5
\][/tex]
So, [tex]\( h(-2) = 5 \)[/tex].
3. Calculate [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]\[
h(4) = \frac{1}{2}(4) - 4 = 2 - 4 = -2
\][/tex]
So, [tex]\( h(4) = -2 \)[/tex].
Summarizing the results:
[tex]\[
\begin{array}{l}
h(-3) = 0 \\
h(-2) = 5 \\
h(4) = -2
\end{array}
\][/tex]
