Question

Find h (-4), h (0), and h (1).

Answer 1

The piecewise function is given below as

`h(x)=\begin{cases}(1)/(2)x-2,x<-2 \\ (x-1)^2-1.-2\leq x\leq1 \\ 2,x>1\end{cases}`

Step 1:

To figure out h(-4),

The function that satisfies this is given below as

`\begin{gathered} (1)/(2)x-2,x<-2 \\ \text{note:} \\ -4<-2 \end{gathered}`

Therefore,

Substitute the value of x=-4

`\begin{gathered} (1)/(2)x-2 \\ h(-4)=(1)/(2)(-4)-2 \\ h(-4)=-2-2 \\ h(-4)=-4 \end{gathered}`

Hence,

h( -4 ) = -4

Step 2:

To figure out h(0)

The function that satisfies this condition is given below as

`\begin{gathered} (x-1)^2-1.-2\leq x\leq1 \\ \text{note:} \\ 0<1 \end{gathered}`

Hence,

Substitue the value of x = 0

`\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}`

Hence,

The value for h( 0 ) = 0

Step 3:

To figure out h( 1)

The function that satisfies this condition is given below as

`(x-1)^2-1.-2\leq x\leq1`

Substitute x= 1 in the equation below

`\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}`

Hence,

The value of h(1) = -1