Solution
Given that function h is defined, for all real numbers as follows
[tex]h(x)=\begin{cases}\frac{1}{2}x^2-5\text{ if x}\ne-2 \\ 1\text{ if x}=-2\end{cases}[/tex]
For h(-5)
[tex]\begin{gathered} I\text{f x}-2 \\ h(x)=\frac{1}{2}x^2-5 \\ h(-5)=\frac{1}{2}(-5)^2-5=\frac{1}{2}(25)-5=12.5-5=7.5 \\ h(-5)=7.5 \end{gathered}[/tex]
Hence, h(-5) is 7.5
For h(-2)
[tex]\begin{gathered} h(-2)\text{ if x}=-2 \\ I\text{f x}=-2,\text{ h\lparen x\rparen}=1 \\ h(-2)=1 \end{gathered}[/tex]
Hence, h(-2) is 1
For h(3)
[tex]\begin{gathered} If\text{ x}\ne-2 \\ h(x)=\frac{1}{2}x^2-5 \\ h(3)=\frac{1}{2}(3)^2-5=\frac{1}{2}(9)-5=4.5-5=-0.5 \\ h(3)=-0.5 \end{gathered}[/tex]
Hence, h(3) is -0.5
