Question

How do you find the critical numbers for `f(x)= x sqrt (2x+1)` to determine the maximum and minimum?

Answer 1

Critical points are `x=-1/2`, where `f'(x)` is not defined and `x=-1/3`, where we have a minima for `f(x)`

Explanation:

We say that `x=a` is a critical point of the function `f(x)` if `f(a)` exists and if either `f'(a)=0` or `f'(a)` does not exist.

Observe that `f(x)=xsqrt(2x+1)` if `x < -1/2`

Let us therefore workout `f'(x)` using product rule.

`f'(x)=x xx1/(2sqrt(2x+1))xx2+1xxsqrt(2x+1)`

= `x/(sqrt(2x+1))+sqrt(2x+1)`

= `(x+2x+1)/(sqrt(2x+1))=(3x+1)/(sqrt(2x+1))`

Now `f'(x)=0`, when `x=-1/3`. Also `f'(x)` is not defined for `x <= -1/2`, hence `x=-1/2` is a critical point.

Further `f''(x)=(sqrt(2x+1)xx3-(3x+1)xx(1/sqrt(2x+1)))/(2x+1)`

= `(6x+3-3x-1)/((2x+1)sqrt(2x+1))=(3x+2)/((2x+1)sqrt(2x+1)`

and at `x=-1/3`, `f''(x) > 0` and hence we have a minima at `x=-1/3`, where `f(x)=-1/(3sqrt3)`.

graph{xsqrt(2x+1) [-4.08, 5.92, -0.86, 4.14]}