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

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Answer 1 · 2017-04-28T13:16:15

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

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]}

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