 # Find the absolute maximum and absolute minimum values of function on the given interval. f(x)=5+54x-2x^3, [0,4] usagirl007A 2021-03-06 Answered
Find the absolute maximum and absolute minimum values of function on the given interval.
$f\left(x\right)=5+54x-2{x}^{3},\left[0,4\right]$
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Step 1
Given:
Given that function $f\left(x\right)=5+54x-2{x}^{3}$
The given interval is $\left[0,4\right]$
To find:
we have to find the absolute maximum and absolute minimum
Step 2
Here $f\left(x\right)=5+54x-2{x}^{3}$ ...(1) and the interval is $\left[0,4\right]$
Differentiate (1) w.r.t.x, we get
$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(5+54x-2{x}^{3}\right)$
${f}^{\prime }\left(x\right)=54-6{x}^{2}$
set ${f}^{\prime }\left(x\right)=0$
$⇒54-6{x}^{2}=0$
$⇒6{x}^{2}=54$
$⇒{x}^{2}=9$
$⇒x=3$ and $x=-3$
but $x=-3\notin \left[0,4\right]$
and it also contain the critical point $x=0$
Step 3
Hence, the critical points is $x=3$ and $x=0$
Now, $f\left(x\right)=5+54x-2{x}^{3}$
For $x=3$
$f\left(3\right)=5+54\left(3\right)-2{\left(3\right)}^{3}=5+162-54$
$=113$
Thus, the absolute value is maximum at $x=3$
Now, $f\left(x\right)=5+54x-2{x}^{3}$
For $x=0$
$f\left(0\right)=5$
Thus, the absolute value is minimum at $x=0$