# Consider the function f(x)=x^4-32x^2+11, -3 <= x <= 9. This function has an absolute minimum value equal to ? and an absolute maximum value equal to ?

Consider the function $f\left(x\right)={x}^{4}-32{x}^{2}+11,-3\le x\le 9$. This function has an absolute minimum value equal to ? and an absolute maximum value equal to ?
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Step 1
Consider the given function,
$f\left(x\right)={x}^{4}-32{x}^{2}+11,-3\le x\le 9$
differentiate the function with respect to x,
${f}^{\prime }\left(x\right)=\frac{d}{dx}\left({x}^{4}-32{x}^{2}+11\right)$
$=\frac{d}{dx}\left({x}^{4}\right)-\frac{d}{dx}\left(32{x}^{2}\right)+\frac{d}{dx}\left(11\right)$
$=4{x}^{3}-64x$
for the maxima and minima, f′(x)=0,
$4{x}^{3}-64x=0$
$4x\left({x}^{2}-16\right)=0$
x=0,-4,4
the values of x lie in the interval $-3\le x\le 9$ is,
x=0,4
Step 2
Now, use the second derivative test so differentiate again,
$f{}^{″}\left(x\right)\frac{d}{dx}\left(4{x}^{3}-64x\right)$
$=12{x}^{2}-64$
at x = 0,
$f{}^{″}\left(x\right)=12{\left(0\right)}^{2}-64$
=-64
f''(x)<0
so, the function has the absolute maximum at x=0,
now, for the absolute maximum value substitute x=0 in the function,
$f\left(0\right)=\left({\left(0\right)}^{4}\right)-32{\left(0\right)}^{2}+11$
=0-0+11
=11
the absolute maximum value of the function is 11.
Step 3
at x=4,
$f{}^{″}\left(x\right)=12{\left(4\right)}^{2}-64$
$=12×16-64$
=128
f''(x)>0
so, the function has the absolute minimum at x=4,
now, for the absolute minimum value substitute x=4 in the function,
$f\left(4\right)={\left(4\right)}^{4}-32{\left(4\right)}^{2}+11$
=256-512+11
=-245
the absolute minimum value of the function is −245.