# Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=4x^3-6x^2-24x+9,[-2,3] absolute minimum value-? absolute maximum value-?

Find the absolute maximum and absolute minimum values of f on the given interval.
$f\left(x\right)=4{x}^{3}-6{x}^{2}-24x+9,\left[-2,3\right]$
absolute minimum value-?
absolute maximum value-?
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dessinemoie
Step 1
We will find the first derivative using the power rule
$f\left(x\right)=4{x}^{3}-6{x}^{2}-24x+9$
${f}^{\prime }\left(x\right)=4\left(3{x}^{2}\right)-6\left(2x\right)-24$
${f}^{\prime }\left(x\right)=12{x}^{2}-12x-24$
Step 2
We find the critical values by solving f'(x)=0
$12{x}^{2}-12x-24=0$
$12\left({x}^{2}-x-2\right)=0$
12(x-2)(x+1)=0
x=2,-1
Then we find the values of the function at the critical points and at the endpoints.
$f\left(x\right)=4{x}^{3}-6{x}^{2}-24x+9$
$f\left(-2\right)=4{\left(-2\right)}^{3}-6{\left(-2\right)}^{2}-24\left(-2\right)+9=1$
$f\left(-1\right)=4{\left(-1\right)}^{3}-6{\left(-1\right)}^{2}-24\left(-1\right)+9=23$ (max)
$f\left(2\right)=4{\left(2\right)}^{3}-6{\left(2\right)}^{2}-24\left(2\right)+9=-31$ (min)
$f\left(3\right)=4{\left(3\right)}^{3}-6{\left(3\right)}^{2}-24\left(3\right)+9=-9$
Result:
Absolute minimum value= -31
Absolute maximum value=23
Jeffrey Jordon