# Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = \ln(x^{2} + 5x + 13), \left[−3, 1\right]

Find the absolute maximum and absolute minimum values of f on the given interval.
$f\left(x\right)=\mathrm{ln}\left({x}^{2}+5x+13\right),\left[-3,1\right]$
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Aamina Herring

Step 1
The given function
$f\left(x\right)=\mathrm{ln}\left({x}^{2}+5x+13\right),\left[-3,1\right]$
Differentiate with respect to x
${f}^{prime}\left(x\right)=\frac{d\left(\mathrm{ln}\left({x}^{2}+5x+13\right)\right)}{d\left({x}^{2}+5x+13\right)}×\frac{d\left({x}^{2}+5x+13\right)}{dx}$
$=\frac{2x+5}{\left({x}^{2}+5x+13\right)}$
Step 2
${f}^{prime}\left(x\right)=0$
$x=\frac{-5}{2}$
So critical points
$x=\frac{-5}{2},-3,1$
$f\left(-2.5\right)=\mathrm{ln}\left({\left(-2.5\right)}^{2}-5×2.5+13\right)=1.9$
$f\left(-3\right)=\mathrm{ln}\left({\left(-3\right)}^{2}-5×3+13\right)=1.95$
$f\left(1\right)=\mathrm{ln}\left({\left(1\right)}^{2}+5×1+13\right)=\mathrm{ln}\left(19\right)=2.94$
Step 3
Absolute maximum value = 2.94
Absolute minimum value = 1.9

Jeffrey Jordon