Question

Is `f(x) =cscx-x` concave or convex at `x=pi/3`?

Answer 1

`f` is convex or concave up at `x=pi/3`.

Explanation:

First, some terminology:

`f` is convex, which is more commonly known as being concave up at `x=a`, if `f''(a)>0`.

`f` is concave, which is more commonly known as being concave down at `x=a`, if `f''(a)<0`.

So, we need to find `f''(pi/3)` when `f(x0=cscx-x`.

It will be helpful to know that:

• `d/dxcscx=-cscxcotx`
• `d/dxcotx=-csc^2x`

Finding the first derivative:

`f(x)=cscx-x`

`f'(x)=-cscxcotx-1`

For the second derivative, we'll need to use the product rule:

`f''(x)=-(d/dxcscx)cotx-cscx(d/dxcotx)`

`f''(x)=-(-cscxcotx)cotx-cscx(-csc^2x)`

`f''(x)=cscxcot^2x+csc^3x`

Evaluating at `x=pi/3`, we should recognize that

• `csc(pi/3)=1/sin(pi/3)=1/(sqrt3/2)=2/sqrt3`
• `cot(pi/3)=cos(pi/3)/sin(pi/3)=(1/2)/(sqrt3/2)=1/sqrt3`

So:

`f''(pi/3)=2/sqrt3(1/sqrt3)^2+(2/sqrt3)^3`

Without having to simplify this, we see that all the terms are positive, so `f''(pi/3)>0` as well. Thus, `f` is convex or concave up at `x=pi/3`.