Question

Find the extrema ; please help the following question ?

Answer 1

For `f(x)=x^5-5x`, we have a relative maxima at `x=-1` and relative minima at `x=1`

Explanation:

Extrema is at a point where `(df)/(dx)=0` and if at that point `(d^2f)/(dx^2)>0`, we have a relative minima and if at that point `(d^2f)/(dx^2)<0`, we have a relative maxima.

Now we are given `f(x)=x^5-5x`, hence `(df)/(dx)=5x^4-5` and is `0`, when `5x^4-5=0` or `x^4-1=0` i.e. at `x=+-1`

We have `(d^2f)/(dx^2)=20x^3`

and at `x=-1` `(d^2f)/(dx^2)=-20<0` and hence, we have a relative maxima and

at `x=1`, `(d^2f)/(dx^2)=20>0` and hence, we have a relative minima.

graph{x^5-5x [-10, 10, -5, 5]}