Question

Suppose that the function h is defined, for all real numbers, as follows.Find h(- 5), h(- 2) and h(3) .h(x)= 1 2 x^ 2 -5&if x=-2\\ 1&if x=-2

Answer 1

Given that function h is defined, for all real numbers as follows

`h(x)=\begin{cases}(1)/(2)x^2-5\text{ if x}\\e-2 \\ 1\text{ if x}=-2\end{cases}`

For h(-5)

`\begin{gathered} I\text{f x}-2 \\ h(x)=(1)/(2)x^2-5 \\ h(-5)=(1)/(2)(-5)^2-5=(1)/(2)(25)-5=12.5-5=7.5 \\ h(-5)=7.5 \end{gathered}`

Hence, h(-5) is 7.5

For h(-2)

`\begin{gathered} h(-2)\text{ if x}=-2 \\ I\text{f x}=-2,\text{ h\lparen x\rparen}=1 \\ h(-2)=1 \end{gathered}`

Hence, h(-2) is 1

For h(3)

`\begin{gathered} If\text{ x}\\e-2 \\ h(x)=(1)/(2)x^2-5 \\ h(3)=(1)/(2)(3)^2-5=(1)/(2)(9)-5=4.5-5=-0.5 \\ h(3)=-0.5 \end{gathered}`

Hence, h(3) is -0.5