# What is the minimum value of g(x) = (x-1)/(x^2+4)? on the interval [-2,2]?

Dec 22, 2016

Minimum value is at $x = 1 - \sqrt{5} \approx \text{-} 1.236$;

$g \left(1 - \sqrt{5}\right) = - \frac{1 + \sqrt{5}}{8} \approx \text{-} 0.405$.

#### Explanation:

On a closed interval, the possible locations for a minimum will be:

• a local minimum inside the interval, or
• the endpoints of the interval.

We therefore compute and compare values for $g \left(x\right)$ at any $x \in \left[\text{-2} , 2\right]$ that makes $g ' \left(x\right) = 0$, as well as at $x = \text{-2}$ and $x = 2$.

First: what is $g ' \left(x\right)$? Using the quotient rule, we get:

$g ' \left(x\right) = \frac{\left(1\right) \left({x}^{2} + 4\right) - \left(x - 1\right) \left(2 x\right)}{{x}^{2} + 4} ^ 2$
$\textcolor{w h i t e}{g ' \left(x\right)} = \frac{{x}^{2} + 4 - 2 {x}^{2} + 2 x}{{x}^{2} + 4} ^ 2$
$\textcolor{w h i t e}{g ' \left(x\right)} = - \frac{{x}^{2} - 2 x - 4}{{x}^{2} + 4} ^ 2$

This will equal zero when the numerator is zero. By the quadratic formula, we get

x^2-2x-4=0" "=>" "x=1+-sqrt 5 approx {"-1.236", 3.236}

Only one of these $x$-values is in $\left[\text{-2} , 2\right]$, and that is $x = 1 - \sqrt{5}$.

Now, we compute:

1. g("-2") = ("-"2-1)/(("-2")^2+4)="-3"/8="-"0.375

2. $g \left(1 - \sqrt{5}\right) = \frac{1 - \sqrt{5} - 1}{{\left(1 - \sqrt{5}\right)}^{2} + 4} = \frac{\text{-} \sqrt{5}}{1 - 2 \sqrt{5} + 5 + 4}$
$\textcolor{w h i t e}{g \left(1 - \sqrt{5}\right)} = - \frac{\sqrt{5}}{10 - 2 \sqrt{5}} = - \frac{\sqrt{5}}{\left(2\right) \left(5 - \sqrt{5}\right)} \cdot \textcolor{b l u e}{\frac{5 + \sqrt{5}}{5 + \sqrt{5}}}$
color(white)(g(1 - sqrt 5)) =-(5+5 sqrt 5)/(2 * (25-5)
$\textcolor{w h i t e}{g \left(1 - \sqrt{5}\right)} = - \frac{5 \left(1 + \sqrt{5}\right)}{40} = - \frac{1 + \sqrt{5}}{8} \approx \text{-} 0.405$

3. $g \left(2\right) = \frac{2 - 1}{{2}^{2} + 4} = \frac{1}{8} = 0.125$

Comparing these three values of $g \left(x\right)$, we see that $g \left(1 - \sqrt{5}\right)$ is the smallest. So $- \frac{1 + \sqrt{5}}{8}$ is our minimum value for $g \left(x\right)$ on $\left[\text{-} 2 , 2\right]$.