Answer:
The piecewise function is given below as
[tex]h(x)=\begin{cases}\frac{1}{2}x-2,x<-2 \\ (x-1)^2-1.-2\leq x\leq1 \\ 2,x>1\end{cases}[/tex]
Step 1:
To figure out h(-4),
The function that satisfies this is given below as
[tex]\begin{gathered} \frac{1}{2}x-2,x<-2 \\ \text{note:} \\ -4<-2 \end{gathered}[/tex]
Therefore,
Substitute the value of x=-4
[tex]\begin{gathered} \frac{1}{2}x-2 \\ h(-4)=\frac{1}{2}(-4)-2 \\ h(-4)=-2-2 \\ h(-4)=-4 \end{gathered}[/tex]
Hence,
h( -4 ) = -4
Step 2:
To figure out h(0)
The function that satisfies this condition is given below as
[tex]\begin{gathered} (x-1)^2-1.-2\leq x\leq1 \\ \text{note:} \\ 0<1 \end{gathered}[/tex]
Hence,
Substitue the value of x = 0
[tex]\begin{gathered} (x-1)^2-1.-2\leq x\leq1 \\ h(0)=(0-1)^2-1 \\ h(0)=(-1)^2-1 \\ h(0)=1-1 \\ h(0)=0 \end{gathered}[/tex]
Hence,
The value for h( 0 ) = 0
Step 3:
To figure out h( 1)
The function that satisfies this condition is given below as
[tex](x-1)^2-1.-2\leq x\leq1[/tex]
Substitute x= 1 in the equation below
[tex]\begin{gathered} (x-1)^2-1.-2\leq x\leq1 \\ h(1)=(1-1)^2-1 \\ h(1)=0-1 \\ h(1)=-1 \end{gathered}[/tex]
Hence,
The value of h(1) = -1
