MHB Mattyk's question at Yahoo Answers regarding local extrema

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The function f(x) = -2x^3 + 21x^2 - 36x + 5 has one local minimum and one local maximum. The local minimum occurs at x = 1 with a value of -12, while the local maximum occurs at x = 6 with a value of 113. Critical values were found by setting the first derivative to zero, yielding x = 1 and x = 6. The second derivative test confirmed the nature of the extrema at these points. This analysis provides a clear understanding of the local extrema for the given cubic function.
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Here is the question:

Calculus homework help?!?

The function f(x) = -2 x^3 + 21 x^2 - 36 x + 5 has one local minimum and one local maximum.
This function has a local minimum at x equals ? with value ?
and a local maximum at x equals ? with value ?

Here is a link to the question:

Calculus homework help?!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Re: mattyk's question as Yahoo! Answers regarding local extrema

Hello mattyk,

We are given the function:

$f(x)=-2x^3+21x^2-36x+5$

To determine the critical values, i.e., the $x$-values at which the local extrema occur, we need to equate the first derivative to zero:

$f'(x)=-6x^2+42x-36=-6(x^2-7x+6)=-6(x-1)(x-6)=0$

Thus, our two critical values are:

$x=1,\,6$

To determine the nature of the extrema at these points, we may look at the sign of the second derivative there:

$f''(x)=-12x+42=-6(2x-7)$

$f''(1)>0$ thus $(1,f(1))=(1,-12)$ is a local minimum.

$f''(6)<0$ thus $(6,f(6))=(6,113)$ is a local maximum.
 
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