Question

Let f(x) = 2x3 – 3x2 – 12x + 15 on (-2,4]. The relative
maximum occurs at x =​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO CALCULATE

The relative maximum for f(x) on the interval ( - 2,4 ]

Where

$\sf{ f(x) = 2 {x}^{3} - 3 {x}^{2} - 12x + 15 \: \: }$

CALCULATION

Here

$\sf{ f(x) = 2 {x}^{3} - 3 {x}^{2} - 12x + 15 \: \: }$

Differentiating both sides with respect to x two times we get

$\sf{f'(x)= \: 6 {x}^{2} - 6x - 12 \: }$

and

$\sf{f''(x)= 12x - 6\: }$

Now for extremum value of f(x) we have

$\sf{f \: '(x)= 0\: }$

$\implies \sf{ 6 {x}^{2} - 6x - 12 = 0\: }$

$\implies \sf{ {x}^{2} - x - 2 = 0\: }$

$\implies \sf{ {x}^{2} -2 x + x- 2 = 0\: }$

$\implies \sf{ x(x - 2) + 1( x- 2) = 0\: }$

$\implies \sf{ (x - 2)( x + 1) = 0\: }$

$\implies \sf{x = - 1 \: \: , \: \: 2 }$

Now

$\sf{f''(2)= 24 - 6 = 18 > 0\: }$

So f(x) has a minimum value at x = 2

Again

$\sf{f''( - 1)= - 12 - 6 = - 18 < 0\: }$

So f(x) has a maximum at x = - 1

Hence the maximum value is

$\sf{ f( - 1) = 2 {( - 1)}^{3} - 3 {( - 1)}^{2} - 12 \times ( - 1) + 15 \: \: }$

$= \sf{ - 2 - 3 + 12 + 15 \: }$

$= \sf{27 - 5 \: }$

$= \sf{22 \: }$

RESULT

Hence the relative maximum for the function

$\sf{ f(x) = 2 {x}^{3} - 3 {x}^{2} - 12x + 15 \: \: }$

on the interval ( - 2,4 ] occurs at x = - 1 and the maximum value is 22

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