Question

How do you find local maximum value of f using the first and second derivative tests: `f(x)= sinx`?

Answer 1

The max occurs at `x=pi/2` where `f(x)=1`

Explanation:

`1.` The 1st derivative test: `f'(x)=cos x`

`cos x=0` at `x= pi/2, (3pi)/2, "etc."`

Make a sign chart for `f'(x)` with a critical point at `pi/2`. To the left of `pi/2`, `cos x` is positive; to the right, it is negative. Therefore, there is a max at `pi/2`

`2.` The 2nd derivative test: `f^('')(x) = - sin (x)`

At the critical point where `x=pi/2`, you have

`-sin(pi/2) = -1 <0`

Since the second derivative is negative at `x=pi/2`, the curve is concave down at that point. Therefore, it is a max.