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Does $f(x)=x^3+2x+\tan x$ have any local maximum or minimum values? Justify your answer.

Sorry I had to ask this question without even showing my steps-I just couldn't get started.

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    I'm going into the e$x$am without a graphic calculator =( Been weaning myself off those calculators.2012-09-28

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The derivative of this function is positive everywhere, so there can be no local extreme values except at endpoints, and there are no endpoints within the domain.

The tangent function is periodic and has vertical asymptotes at odd-integer multiples of $\pi/2$. So this function is increasing on each of the intervals bounded by two successive asymptotes. It's not increasing on its domain as a whole, however, since it goes down to $-\infty$ at the left end of each of those intervals and up to $+\infty$ at the right end.

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    Typo: I wrote "the positive number $s$"; I meant the positive number $2$.2012-09-28
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Suppose your functions doesn't have local maximums/minimums. Then it will be always monotonic. At least non-decreasing.

if you prove it's non-decreasing (the signs tells that), then it's trivial it has no local maximums/minimums because it should be a point that's greater than other points to the right.


An easy way is to prove that three components of it are non-decreasing, so the sum should be non-decreasing too.

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    Good point. Didn't realize the non-continous nature of tan x2012-09-28
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Find the second derivative. Allow it to be equal to zero and solve for $x$, then substitute those values into the second derivative. If you get a positive answer then a local minima exist likewise a negative implies a local maxima

Here is the source.

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    Damn right. Second derivative test! Let me try that2012-09-28