Question

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

Answer 1

`f (x)=csc x- sin x` is CONVEX at `x=pi/3`

Explanation:

From the given equation
`f (x)=csc x- sin x`
We determine `f'' (x)`, the second derivative of `f(x)`

after which , we test if `f'' (pi/3)>0`, if True then CONVEX.
If `f'' (pi/3)<0`, if True then CONCAVE

Let us determine first `f''(x)` by obtaining first derivative `f' (x)`

`f (x)=csc x- sin x`

`f' (x)=d/dxf(x)=d/dx(csc x)-d/dx(sin x)`
`f' (x)=-csc x * cot x - cos x" "`the first derivative

Let us determine now the second derivative `f'' (x)`

`f'' (x)=d/dxf' (x)=d/dx(-csc x * cot x) - d/dx(cos x)" "`

`f'' (x)=-[cot x*d/dx(csc x)+csc x*d/dx(cot x)]-(-sin x)" "`

`f'' (x)=-[cot x*(-csc x cot x)+csc x(-csc^2 x)]+sin x`

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

Substitute now `x=pi/3` in `f'' (x)=csc x cot^2 x+ csc^3 x+sin x`

`f'' (x)=csc x cot^2 x+ csc^3 x+sin x`
`f'' (pi/3)=csc (pi/3) cot^2 (pi/3)+ csc^3 (pi/3)+sin (pi/3)`

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

`f'' (pi/3)>0` therefore CONVEX

God bless....I hope the explanation is useful.

Answer 2

`f(x)` is convex at `x=pi/3`

Explanation:

A concave function is a function in which no line segment joining two points on its graph lies above the graph at any point.

A convex function, on the other hand, is a function in which no line segment joining two points on the graph lies below the graph at any point.

It means that, if `f(x)` is more than the average of `f(x+-lambda)` than the function is concave and if `f(x)` is less than the average of `f(x+-lambda)` than the function is convex.

Let us first calculate `f(pi/3)=csc(pi/3)-sin(pi/3)=2/sqrt3-sqrt3/2=(4-3)/(2sqrt3)=1/2sqrt3=sqrt3/6=0.2887`

As `pi/3=60*o`, let us also calculate `f(55^o)` and `f(65^o)`

`f(55^o)=csc55^o-sin55^o=1.22077-0.81915=0.40162` and

`f(65^o)=csc65^o-sin65^o=1.10338-0.90631=0.19707`

The average of two is `(0.40162+0.19707)/2=0.29935`

As `f(pi/3)<0.29935`, `f(x)` is convex at `x=pi/3`