Question

In which interval the function `f(x)=sqrt(x^2+8)-x` is a decreasing function?

Answer 1

`f(x)=sqrt(x^2+8)-x` is a decreasing function for all values of `x`

Explanation:

Intervals of increasing of a function are those where `f'(x)>0` and intervals of decreasing for a function are those where `f'(x)<0`.

Here `f(x)=sqrt(x^2+8)-x`

and `f'(x)=(df)/(dx)=1/(2sqrt(x^2+8))xx2x-1`

`=x/sqrt(x^2+8)-1`

`=(x-sqrt(x^2+8))/sqrt(x^2+8)`

It is apparent that as `x^2+8` is always positive and `sqrt(x^2+8)>x`,

`x-sqrt(x^2+8)` is negative for all values of `x`

i.e. for all `x` `f'(x)<0`.

Hence, `f(x)=sqrt(x^2+8)-x` is a decreasing function for all values of `x`

graph{sqrt(x^2+8)-x [-8.125, 11.875, -3.28, 6.72]}