# Find if f(x)=3x^{4}-4x^{3} has any absolute extreme value/values in (-2,

Find if $f\left(x\right)=3{x}^{4}-4{x}^{3}$ has any absolute extreme value/values in $\left(-2,+\mathrm{\infty }\right)$.
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Step 1
Given
$f\left(x\right)=3{x}^{4}-4{x}^{3}$
To find
Absolute maximum and absolute minimum values.
Step 2
$f\left(x\right)=3{x}^{4}-4{x}^{3}$
${f}^{\prime }\left(x\right)=12{x}^{3}-12{x}^{2}$
f'(x)=0
$12{x}^{3}-12{x}^{2}=0$
$12{x}^{2}\left(x-1\right)=0$
${x}^{2}=0,\left(x-1\right)=0$
x=0,0,1
Critical points=0,1
f(0)=0 , f(1)=-1.
$f\left(-2\right)=3{\left(-2\right)}^{4}-4{\left(-2\right)}^{3}=3×16+4\left(8\right)=48+32=80$.
The absolute maximum value is 80 at -2 and minimum value is -1 at 1.