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

Apr 28, 2017

Critical points are $x = - \frac{1}{2}$, where $f ' \left(x\right)$ is not defined and $x = - \frac{1}{3}$, where we have a minima for $f \left(x\right)$

#### Explanation:

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

Observe that $f \left(x\right) = x \sqrt{2 x + 1}$ if $x < - \frac{1}{2}$

Let us therefore workout $f ' \left(x\right)$ using product rule.

$f ' \left(x\right) = x \times \frac{1}{2 \sqrt{2 x + 1}} \times 2 + 1 \times \sqrt{2 x + 1}$

= $\frac{x}{\sqrt{2 x + 1}} + \sqrt{2 x + 1}$

= $\frac{x + 2 x + 1}{\sqrt{2 x + 1}} = \frac{3 x + 1}{\sqrt{2 x + 1}}$

Now $f ' \left(x\right) = 0$, when $x = - \frac{1}{3}$. Also $f ' \left(x\right)$ is not defined for $x \le - \frac{1}{2}$, hence $x = - \frac{1}{2}$ is a critical point.

Further $f ' ' \left(x\right) = \frac{\sqrt{2 x + 1} \times 3 - \left(3 x + 1\right) \times \left(\frac{1}{\sqrt{2 x + 1}}\right)}{2 x + 1}$

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

and at $x = - \frac{1}{3}$, $f ' ' \left(x\right) > 0$ and hence we have a minima at $x = - \frac{1}{3}$, where $f \left(x\right) = - \frac{1}{3 \sqrt{3}}$.

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