Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 52

Determine the absolute maximum and absolute minimum values of $\displaystyle f(x) = \frac{x^2-4}{x^2+4}$ on the interval $[-4,4]$.

Taking the derivative of $f(x)$ using Quotient Rule we have,

$\begin{equation} \begin{aligned} f'(x) &= \frac{\left( x^2 + 4 \right)(2x) - \left( x^2 - 4\right) (2x)}{\left( x^2 + 4\right)^2}\\ \\ f'(x) &= \frac{\cancel{2x^3} + 8x - \cancel{2x^3} + 8x }{\left( x^2 + 4 \right)^2}\\ \\ f'(x) &= \frac{16x}{\left( x^2 + 4 \right)^2} \end{aligned} \end{equation}$

Solving for critical numbers, when $f'(x) = 0$


$\begin{equation} \begin{aligned} 0 &= \frac{16x}{\left( x^2 + 4 \right)^2}\\ \\ 0 &= 16 x\\ \\ x &= 0 \end{aligned} \end{equation}$



We have either absolute maximum and minimum values at $x = 0$
So,

$\begin{equation} \begin{aligned} f(0) &= \frac{0^2 - 4 }{0^2 + 4}\\ \\ f(0) &= -1 \end{aligned} \end{equation}$

Evaluating $f(x)$ at end points $x = -4$ and $x = 4$

$\begin{equation} \begin{aligned} \text{when } x &= -4,\\ \\ f(-4) &= \frac{(-4)^2 -4 }{(-4)^2 + 4}\\ \\ f(-4) &= \frac{3}{5}\\ \\ \\ \\ \text{when } x &= 4,\\ \\ f(4) &= \frac{(4)^2 - 4}{(4)^2 + 4}\\ \\ f(4) &= \frac{3}{5} \end{aligned} \end{equation}$

Therefore, we have absolute maximum value at $\displaystyle f(-4) = f(4) = \frac{3}{5}$ and the absolute minimum value at $f(0) = -1$ on the interval $[-4,4]$