# What are the critical points of f(x) = 1/sqrt(x+x^2-3)?

Dec 7, 2015

The critical points for $f \left(x\right)$ are

$\left\{- \frac{1}{2} , \frac{- 1 + \sqrt{13}}{2} , \frac{- 1 - \sqrt{13}}{2}\right\}$

#### Explanation:

The critical points of a function $f \left(x\right)$ are the points $c$ at which $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ is undefined.
Thus first we find the derivative of the function, and then set it equal to zero.

To find the derivative we apply the quotient rule, the chain rule, and the power rule:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \frac{1}{\sqrt{x + {x}^{2} - 3}}$

$= - \frac{\frac{1}{2} {\left(x + {x}^{2} - 3\right)}^{- \frac{1}{2}} \left(2 x + 1\right)}{x + {x}^{2} - 3}$

$= - \frac{2 x + 1}{2 {\left(x + {x}^{2} - 3\right)}^{\frac{3}{2}}}$

Now, to find the critical points, we want to find all points where $f '$ is $0$ or undefined.

Because $f ' \left(x\right) = 0 \iff 2 x + 1 = 0$
we can solve for $x$ to find a critical point at $- \frac{1}{2}$.

As $f ' \left(x\right)$ is undefined if and only if
$2 {\left(x + {x}^{2} - 3\right)}^{\frac{3}{2}} = 0 \iff {x}^{2} + x - 3 = 0$
we can use the quadratic formula to find critical points at
$\frac{- 1 \pm \sqrt{13}}{2}$

Thus the critical points for $f \left(x\right)$ are

$\left\{- \frac{1}{2} , \frac{- 1 + \sqrt{13}}{2} , \frac{- 1 - \sqrt{13}}{2}\right\}$