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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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    Since you tagged the question "calculus", can you at least name the _general_ calculus-based technique for investigating local maxima and minima?2012-09-28
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    Couldn't get started? You don't know any methods at all for finding local maxima and/or minima? You've tagged this Calculus --- surely you've been exposed to at least *one* method.2012-09-28
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    Well ok... $f'(x)=3x^2+2+\sec^2 x$ Now I'm stuck after letting it be $3x^2+2+\sec^2 x= 0$2012-09-28
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    Did you plot the function and then use the methods you are learning in Calculus to at least get you going? Start with a plot using wolframalpha.2012-09-28
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    I'm going into the exam 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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    $3x^2+2+\sec^2 x$ is positive everywhere? How can you tell?2012-09-28
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    @SingaporeanDude. - every element in the sum is positive everywhere. note anything of the form $y^2$ is positive everywhere2012-09-28
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    $3x^2$ is non-negative because it's the product of a square and a positive number. $\sec^2 x$ is always positive because it's a square and the secant function has no real zeros. Then you're adding the positive number $s$ to it.2012-09-28
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    Gosh I am so blind. Thanks a bunch. But can it be equals to 0? Never mind, just saw the 22012-09-28
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    No. $3x^2 \geq 0, 2 > 0, \sec^2 x > 0$. So $f'$ is even always greater than 2.2012-09-28
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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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    It's non-decreasing on each connected component of its domain separately, but it's not non-decreasing on its domain as a whole.2012-09-28
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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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    But there are no points where the first derivative is $0$.2012-09-28
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    Damn right. Second derivative test! Let me try that2012-09-28