# How do you find the absolute maximum and absolute minimum values of f on the given interval f(x) = ln(x^2 + 3x + 9) and [-2, 2]?

##### 1 Answer
Apr 14, 2018
• Minimum : $\ln \left(\frac{27}{4}\right)$ at $x = - \frac{3}{2}$
• Maximum : $\ln \left(19\right)$ at $x = 2$

#### Explanation:

Given

$f \left(x\right) = \ln \left({x}^{2} + 3 x + 9\right)$

we have

${f}^{'} \left(x\right) = \frac{2 x + 3}{{x}^{2} + 3 x + 9}$

and

${f}^{' '} \left(x\right) = \frac{\left({x}^{2} + 3 x + 9\right) \frac{d}{\mathrm{dx}} \left(2 x + 3\right) - \left(2 x + 3\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 3 x + 9\right)}{{x}^{2} + 3 x + 9} ^ 2$
$q \quad = \frac{\left({x}^{2} + 3 x + 9\right) \cdot 2 - {\left(2 x + 3\right)}^{2}}{{x}^{2} + 3 x + 9} ^ 2$
$q \quad = \frac{2 {x}^{2} + 6 x + 18 - \left(4 {x}^{2} + 12 x + 9\right)}{{x}^{2} + 3 x + 9} ^ 2$
$q \quad = \frac{9 - 2 {x}^{2} - 6 x}{{x}^{2} + 3 x + 9} ^ 2$

For a local extremum, we have

${f}^{'} \left(x\right) = 0 \implies 2 x + 3 = 0 \implies x = - \frac{3}{2}$

${f}^{' '} \left(- \frac{3}{2}\right) > 0$

So $x = - \frac{3}{2}$ is a local minimum. The minimum value is

$f \left(- \frac{3}{2}\right) = \ln \left({\left(- \frac{3}{2}\right)}^{2} + 3 \left(- \frac{3}{2}\right) + 9\right) = \ln \left(\frac{27}{4}\right)$

Since $x = - \frac{3}{2}$ is the only local extremum in $\left[- 2 , 2\right]$, the maximum is reached at an endpoint.

Now,

$f \left(- 2\right) = \ln \left({\left(- 2\right)}^{2} + 3 \left(- 2\right) + 9\right) = \ln \left(7\right)$

and

$f \left(2\right) = \ln \left({2}^{2} + 3 \cdot 2 + 9\right) = \ln \left(19\right)$

So the global maximum is $\ln \left(19\right)$ at $x = 2$