# How do use the first derivative test to determine the local extrema (x)/[(x^2) +3]?

Refer to explanation

#### Explanation:

The First Derivative Test states

Suppose $f \left(x\right)$ is continuous at a critical point ${x}_{o}$.

If $f ' \left(x\right) > 0$ on an open interval extending left from ${x}_{o}$ and $f ' \left(x\right) < 0$ on an open interval extending right from $x o$, then $f$ has a relative maximum at ${x}_{o}$.
If $f ' \left(x\right) < 0$ on an open interval extending left from ${x}_{o}$ and $f ' \left(x\right) > 0$ on an open interval extending right from ${x}_{o}$, then $f$ has a relative minimum at ${x}_{o}$.
If $f \left(x\right)$ has the same sign on both an open interval extending left from x_ and an open interval extending right from ${x}_{o}$, then $f$ does not have a relative extremum at ${x}_{o}$.

Now we find the first derivative of $f \left(x\right) = \frac{x}{{x}^{2} + 3}$ hence we have that

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

Hence the first derivative nullifies at ${x}_{o} = \sqrt{3}$ and ${x}_{o} = - \sqrt{3}$

According to the first derivative test f has

• local maximum at $x = \sqrt{3}$ with $f \left(\sqrt{3}\right) = \frac{1}{2 \sqrt{3}}$
• local minimum at $x = - \sqrt{3}$with $f \left(- \sqrt{3}\right) = - \frac{1}{2 \sqrt{3}}$