Question

Is `f(x)=sinx` concave or convex at `x=pi/5`?

Answer 1

`f(x)=sinx` is concave at `pi/5`

Explanation:

`f(x)` is concave at `x_0` if `f(x_0)>1/2(f(x_0-h)+f(x_0+h))` and

`f(x)` is convex at `x_0` if `f(x_0)<1/2(f(x_0-h)+f(x_0+h))`,

where `(x_0-h)` and `(x_0+h)` are two values around `x_0`

As `pi/5=36^o`, let us calculate `sinx` at `{x=35^o,36^o,37^o}`

`sin35^o=0.5735764`

`sin36^o=0.5877853`

`sin37^o=0.6018150`

Now as `1/2(sin35^o+sin37^o)=1/2(0.5735764+0.6018150)`

= `1/2xx1.1753914=0.5876957< sin36^o`

`f(x)=sinx` is concave at `pi/5`

Further if `f(x)` can be differentiated twice at `x=x_0` and `f''(x_0)<0`, it is concave and if `f''(x_0)>0`, it is convex.

As `f(x)=sinx`, `f'(x)=cosx` and `f''(x)=-sinx` and

`f''(36^o)=-sin36^o=-0.5877853` and hence `f(x)=sinx` is concave at `pi/5`.