# Question #826c5

Jan 30, 2017

The critical points are

$A \left(\sqrt{3} , \frac{\sqrt{3}}{6}\right)$ and $B \left(- \sqrt{3} , - \frac{\sqrt{3}}{6}\right)$.

#### Explanation:

First, we will take the derivative of $F$:

$\frac{\mathrm{dF}}{\mathrm{dt}} = \frac{{t}^{2} + 3 - 2 {t}^{2}}{{t}^{2} + 3} ^ 2 = \frac{- {t}^{2} + 3}{{t}^{2} + 3} ^ 2$

We used the quotient rule: $\left(\frac{a}{b}\right) ' = \frac{a ' b - a b '}{b} ^ 2$.

Now, we must find the values where the derivative is either zero, or undefined. Those will be the critical points. Since the denominator is always positive, it can never be zero, so $F '$ is always defined. However, it can be zero, when the numerator is zero:

$F ' \left(t\right) = 0 \implies t = \pm \sqrt{3}$.

Therefore, the critical points are

$A \left(\sqrt{3} , \frac{\sqrt{3}}{6}\right)$ and $B \left(- \sqrt{3} , - \frac{\sqrt{3}}{6}\right)$.

Extra note:

If we so wanted, we could then find where $F '$ is positive or negative to determine if $F$ is increasing or decreasing. (and since it's differentiable, we know that it's continuous, so we are sure about that last statement)