What are the local extrema of $f (x) = x^{3} - 6 x^{2} + 15$ $f(x)= x^3-6x^2+15$ , if any?

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Answer 1 · 2015-12-03T01:16:39

$(0, 15), (4, - 17)$ $(0,15),(4,-17)$

A local extremum, or a relative minimum or maximum, will occur when the derivative of a function is $0$ $0$ .

So, if we find $f'(x)$ , we can set it equal to $0$ .

$f'(x)=3x^2-12x$

Set it equal to $0$ .

$3x^2-12x=0$

$x(3x-12)=0$

Set each part equal to $0$ .

${(x=0),(3x-12=0rarrx=4):}$

The extrema occur at $(0,15)$ and $(4,-17)$ .

Look at them on a graph:

graph{x^3-6x^2+15 [-42.66, 49.75, -21.7, 24.54]}

The extrema, or changes in direction, are at $(0,15)$ and $(4,-17)$ .

What are the local extrema of f(x)= x^3-6x^2+15f(x)=x3−6x2+15f(x)= x^3-6x^2+15, if any?