How do you find the exact relative maximum and minimum of the polynomial function of $f(x) = –2x^3 + 6x^2 + 18x –18$ ?

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Answer 1 · 2017-08-15T06:50:59

The relative maximum is $=(3,36)$ and the relative minimum is $=(-1,-28)$

The polynomial function is

$f(x)=-2x^3+6x^2+18x-18$

This function is defined, continuous and derivable on $RR$

The first derivative is

$f'(x)=-6x^2+12x+18$

The critical points are when $f'(x)=0$

$-6(x^2-2x-3)=0$

$(x+1)(x-3)=0$

$f'(x)=-(x+1)(x-3)$

Therefore,

$x=-1$ and $x=3$

We can make a variation chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-1$ $color(white)(aaaa)$ $3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+1$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f'(x)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaa)$ $↘$ $color(white)(aaaa)$ $↗$ $color(white)(aaa)$ $↘$

The second derivative is

$f''(x)=-12x+12$

$f''(-1)=12+12=24$ , $=>$ , this is a relative minimum as $f''(-1)>0$

$f''(3)=-12*3+12=-24$ , $=>$ , this is a relative maximum as $f''(-1)<0$

graph{-2x^3+6x^2+18x-18 [-74.1, 74.1, -37.03, 37]}

How do you find the exact relative maximum and minimum of the polynomial function of f(x) = –2x^3 + 6x^2 + 18x –18 f(x) = –2x^3 + 6x^2 + 18x –18 ?