# Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=5+54x-2x^3, x in[0,4]

Find the absolute maximum and absolute minimum values of f on the given interval.
$f\left(x\right)=5+54x-2{x}^{3},x\in \left[0,4\right]$
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smallq9
Given,
$f\left(x\right)=5+54x-2{x}^{3},x\in \left[0,4\right]$
Absolute maximum or absolute minimum values of a function on the given interval exists at a point at which its first derivative is zero or at the end points of the given interval.
So differentiating given function with respect to x, we get
${f}^{\prime }\left(x\right)=0+54\left(1\right)-2\left(3{x}^{2}\right)$
$⇒{f}^{\prime }\left(x\right)=54-6{x}^{2}$
Now f'(x)=0 gives
$54-6{x}^{2}=0$
$⇒6{x}^{2}=54$
$⇒{x}^{2}=9$
$⇒x=±3$
But $-3\notin \left[0,4\right]$, therefore
x=3
Step 2
Now,
$f\left(0\right)=5+54\left(0\right)-2{\left(0\right)}^{3}=5$
$f\left(3\right)=5+54\left(3\right)-2{\left(3\right)}^{3}=113$
$f\left(4\right)=5+54\left(4\right)-2{\left(4\right)}^{3}=93$
Therefore absolute minimum value is 5 occurs at the point x = 0 and absolute maximum value is 113 occur at the point
x =3.