# What are the absolute extrema of f(x)=x / (x^2 -6) in[3,7]?

Nov 7, 2017

The absolute extrema can either occur on the boundaries, on local extrema, or undefined points.

Let us find the values of $f \left(x\right)$ on the boundaries $x = 3$ and $x = 7$. This gives us $f \left(3\right) = 1$ and $f \left(7\right) = \frac{7}{43}$.

Then, find the local extrema by the derivative. The derivative of $f \left(x\right) = \frac{x}{{x}^{2} - 6}$ can be found using the quotient rule: $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} v - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$ where $u = x$ and $v = {x}^{2} - 6$.

Thus, $f ' \left(x\right) = - \frac{{x}^{2} + 6}{{x}^{2} - 6} ^ 2$. The local extrema occurs when $f ' \left(x\right) = 0$, but nowhere in $x \in \left[3 , 7\right]$ is $f ' \left(x\right) = 0$.

Then, find any undefined points. However, for all $x \in \left[3 , 7\right]$, $f \left(x\right)$ is defined.

Therefore, it means that the absolute maximum is $\left(3 , 2\right)$ and the absolute minimum is $\left(7 , \frac{7}{43}\right)$.