Find the absolute max and min values at the indicated interval f(x)=2x^3-x^2-4x+10 ,[-1,0]

Jerold 2021-02-12 Answered
Find the absolute max and min values at the indicated interval
$f\left(x\right)=2{x}^{3}-{x}^{2}-4x+10,\left[-1,0\right]$
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Expert Answer

Roosevelt Houghton
Answered 2021-02-13 Author has 106 answers
Given,
$f\left(x\right)=2{x}^{3}-{x}^{2}-4x+10,\left[-1,0\right]$
Absolute maximum and absolute minimum values of a function exist at a point where its first derivative is zero or at the end points of the given interval.
So first we find the derivative of the given function:
${f}^{\prime }\left(x\right)=2\cdot 3{x}^{2}-2x-4+0$
$⇒{f}^{\prime }\left(x\right)=6{x}^{2}-2x-4$
Now f'(x)=0
$⇒6{x}^{2}-2x-4=0$
$⇒6{x}^{2}-6x+4x-4=0$
$⇒6x\left(x-1\right)+4\left(x-1\right)=0$
$⇒\left(x-1\right)\left(6x+4\right)=0$
$⇒x-1=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}6x+4=0$
$⇒x=1\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=-\frac{2}{3}$
Now computing the value of the function at these points and at the end points of the interval:
at $x=-1,f\left(-1\right)=f\left(x\right)=2{\left(-1\right)}^{3}-{\left(-1\right)}^{2}-4\left(-1\right)+10=11$
at $x=0,f\left(0\right)=f\left(x\right)=2{\left(0\right)}^{3}-{\left(0\right)}^{2}-4\left(0\right)+10=10$
at $x=-\frac{2}{3},f\left(-\frac{2}{3}\right)=f\left(x\right)=2{\left(-\frac{2}{3}\right)}^{3}-{\left(-\frac{2}{3}\right)}^{2}-4\left(-\frac{2}{3}\right)+10=12.8148$
We didn't find the value of f(1), because 1 does not belongs to the interval $\left[-1,0\right]$.
Hence absolute maximum value is 12.8148 and absolute minimum value is 10.
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