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For which x does the function $f(x) = x^3-6x^2-5x+5$ assume its maximum value on the interval $[-5,5]$?

The critical points for this function are $\frac{12 + \sqrt{204}}{6}$ and $\frac{12 - \sqrt{204}}{6}$. The end points are -5 and 5.

The maximum value is solved to be for $\frac{12 - \sqrt{204}}{6}\approx-0.38$.

However when I plug in for $x=0$, the value is larger than that at $-0.38$.

Why?

rae306
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Minu
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  • Perhaps you found a minimal value. When you find critical values they can be maximum or minimum values. Also keep in mind that these minimum and maximum values are not always global maximum/minimum values rather local maximum/minimum values. – ILikeMath Jul 05 '14 at 06:32
  • Use http://en.wikipedia.org/wiki/Second_derivative_test also divide $f(x)$ by $f'(x)$ to express $f(x)$ as linear function of $x$ for the ease of calculation – lab bhattacharjee Jul 05 '14 at 06:38

2 Answers2

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On the interval $[-5,5]$ the function's local maximum value is $y=5.98$ when $x=-0.38$. The other value you found represents the local minimum.

After you have found the critical points with the derivative, you always have to check your answer with a plot of the function.

Plot

rae306
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Looks like an arithmetic error to me. I get a value larger than $5$ when I plug in $x = -0.38$.