Question

Is `f(x)=cosx` concave or convex at `x=pi/2`?

Answer 1

`x=pi/2` corresponds to a transition between convex down and convex up.

Explanation:

`f(x) = cosx`

Differentiating wrt `x` we get
`f(x) = cosx => f'(x) = -sinx`

When `x=pi/2 => f'(pi/2)=-sin(pi/2) = -1`

Now `{ (f'(x)<0, => f(x) " is decreasing"), (f'(x)=0, => f(x) " is stationary"), (f'(x)>0, => f(x) " is increasing") :}`

So we know that `f(x)` is decreasing at `x=pi/2`, we must check the 2nd derivative for the convexity

`f'(x) = -sinx => f''(x)=-cosx`

When `x=pi/2 => f''(pi/2)=-cos(pi/2) = 0`

Now `{ (f''(x)<0, => f(x) " is concave"), (f''(x)=0, => f(x) " is transition point"), (f''(x)>0, => f(x) " is convex") :}`

Confirming that `x=pi/2` corresponds to a transition between convex down and convex up.

The following is a plot of `color(red)(f(x))`, `color(blue)(f'(x))` and `color(green)(f''(x))`. As you can see when `x=pi/2` then `color(red)(f(x))` is decreasing and `color(blue)(f'(x)) < 0`, and `color(green)(f''(x)) = 0`