Examining a Curve

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Jason76

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Oct 19, 2012
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All of this is wrong. Here is the actual question: "Find the local maximum and minimum values of f using both the First and Second Derivative Tests." Please look thru this to see what is mistaken.

\(\displaystyle f(x) = -8x^{3} + 12x^{2} + 8\)

\(\displaystyle f'(x) = -24x^{2} + 24x\)

\(\displaystyle f''(x) = -48x + 24\)

Find Critical Numbers:

\(\displaystyle f'(x) = -24x^{2} + 24x\)

\(\displaystyle -24x^{2} + 24x = 0\)

\(\displaystyle -24x^{2} = - 24\)

\(\displaystyle x^{2} = 1\)

\(\displaystyle \sqrt{x^{2}} = \sqrt{1}\)

\(\displaystyle x = \pm \sqrt{1}\)

\(\displaystyle x = -1\)

\(\displaystyle x = 1\)

Find Min Max

\(\displaystyle f(x) = -8x^{3} + 12x^{2} + 8\)

\(\displaystyle f(-1) = -8(-1)^{3} + 12(-1)^{2} + 8 = 28\)

\(\displaystyle f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12\)
 
\(\displaystyle f(x) = -8x^{3} + 12x^{2} + 8\)

\(\displaystyle f'(x) = -24x^{2} + 24x\)

\(\displaystyle f''(x) = -48x + 24\)

Find Critical Numbers:

\(\displaystyle f'(x) = -24x^{2} + 24x\)

\(\displaystyle -24x^{2} + 24x = 0\)

\(\displaystyle 24x(-x + 1) = 0\) ;) a different approach

\(\displaystyle 24x = 0\)

\(\displaystyle x = 0\)

\(\displaystyle (-x + 1) = 0\)

\(\displaystyle -x = -1\)

\(\displaystyle x = 1\) and \(\displaystyle x = 0\) (Critical Number)


Find Min Max

\(\displaystyle f(x) = -8x^{3} + 12x^{2} + 8\)

\(\displaystyle f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12\) - Max

\(\displaystyle f(0) = -8(0)^{3} + 12(0)^{2} + 8 = 8\) - Min
 
Last edited:
Computer said this problem was right. But still problems on other posts.
 
Jason76 said:
Find Min Max

\(\displaystyle f(x) = -8x^{3} + 12x^{2} + 8\)

\(\displaystyle f(1) = -8(1)^{3} + 12(1)^{2} + 8 = 12\) - Max

\(\displaystyle f(0) = -8(0)^{3} + 12(0)^{2} + 8 = 8\) - Min
Click to expand...

You did not justify here how you know that those values are the max and the min, respectively.
 
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