Question

How we do this one last tutor got it wrong

Answer 1

First, we have to find the derivative

`\begin{gathered} h^(\theta)=3cos((\theta)/(2)) \\ \\ h^(\prime)(\theta)=3(d)/(d\theta)(cos((\theta)/(2))) \\ h^(\prime)(\theta)=3\text{ \lparen - }\sin((\theta)/(2))(d)/(d\theta)((\theta)/(2))) \\ \\ h^(\prime)(\theta)=3(\text{ -}sin((\theta)/(2))((1)/(2)) \\ \\ h^(\prime)(\theta)=\text{ -}(3)/(2)sin((\theta)/(2)) \end{gathered}`
`\begin{gathered} \text{ - }(3)/(2)sin((\theta)/(2))=0 \\ sin((\theta)/(2))=0 \\ \sin^(-1)(0( \\ sin0º=0 \\ andsin180º=0 \\ \\ \\ \end{gathered}`

now he solve for h(theta)

`\begin{gathered} h(\theta)=3cos((\theta)/(2)) \\ \\ h(\text{ -}2\pi)=3cos(\text{ -}(2\pi)/(2))=3cos(\text{ -}\pi)=3(\text{ -1\rparen= -3} \\ h(0)=3cos((0)/(2))=3cos0=3(1)=3 \end{gathered}`