ankarskogC

2021-02-09

Find the absolute maximum and absolute minimum values of the function:
$f\left(x\right)=5{x}^{7}-7{x}^{5}-7$ on the interval $\left[-3,4\right]$

Velsenw

Step 1
We find the critical points by solving ${f}^{\prime }\left(x\right)=0$
Use power rule to find ${f}^{\prime }\left(x\right)$
$f\left(x\right)=5{x}^{7}-7{x}^{5}-7$
${f}^{\prime }\left(x\right)=35{x}^{6}-35{x}^{4}$
$35{x}^{6}-35{x}^{4}=0$
$35{x}^{4}\left({x}^{2}-1\right)=0$
$35{x}^{4}\left(x+1\right)\left(x-1\right)=0$
$x=0,-1,1$
Step 2
Then we find the f(x) values at the endpoints and at the critical points
$f\left(-3\right)=5{\left(-3\right)}^{7}-7{\left(-3\right)}^{5}-7=-9241$ (min)
$f\left(4\right)=5{\left(4\right)}^{7}-7{\left(4\right)}^{5}-7=74745$ (max)
$f\left(0\right)=5{\left(0\right)}^{7}-7{\left(0\right)}^{5}-7=-7$
$f\left(-1\right)=5{\left(-1\right)}^{7}-7{\left(-1\right)}^{5}-7=-5$
$f\left(1\right)=5{\left(1\right)}^{7}-7{\left(1\right)}^{5}-7=-9$
Result: Absolute maximum value= 74745 at $x=4$
Absolute minimum value= -9241 at $x=-3$

Jeffrey Jordon