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Find the maximum and minimum values of: $$ 4 \sin x + \frac{9}{1+\sin x}$$ For $ 0 \le x \le \pi $.

What I have done:

$$ \frac{dy}{dx} = 4\cos x-\frac{9\cos x}{(1+\sin x)^2}$$

After equating the above to $0$, I found that $ x=\pi/6 $.

I do not know what to do next to find the maximum and minimum values. Any help would be appreciated.

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The extreme values must occur at critical points or at endpoints of the interval. Calculate $f(\pi/6)$, $f(0)$, and $f(\pi)$, and compare the values. – Brian M. Scott Jan 8 '13 at 18:19
Draw a picture first. The constraint $x \in [0,\pi]$ may be important, in which case you cannot just look for zeros of the derivative. – copper.hat Jan 8 '13 at 18:19
If not sure on such questions it can be very useful to sketch a graph. – Mark Bennet Jan 8 '13 at 18:19
@BrianM.Scott - you are assuming the function is continuous, which should probably be mentioned given the form it takes (ie note that discontinuities are outside the interval) – Mark Bennet Jan 8 '13 at 18:22
@Mark: Not to worry: I was actually more amused than annoyed. This might have been better conveyed by ‘No, I know that the function is continuous, and I’m assuming (possibly incorrectly) that the OP will check this’! – Brian M. Scott Jan 8 '13 at 18:37
up vote 1 down vote accepted

First, notice that $\dfrac{\pi}{2}$ and $\dfrac{5\pi}{6}$ work as well. You can then use The Extreme Value Theorem. You merely plug in the critical values and the endpoints. The largest is the max, the smallest the min.

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With apologies to Magritte, this is not an answer, but doesn't fit in the comments.

enter image description here

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Thank you for the replies and yes I should have drawn a graph. But is there any method to work out the minimum and maximum values without the use of a graph? – neverloggedin Jan 8 '13 at 18:24
See Brian's comment above. Either the constraints are relevant or not at a solution. If not, using the derivative can characterize a solution, but if a constraint is active, care must be taken. A trivial example is to maximize $x^3$ subject to $|x|=1$. Clearly the solution is $x=1$, but the derivative is of no help here. – copper.hat Jan 8 '13 at 18:25
You remind me of my all-time favorite Usenet sig; see the bottom of this post. – Brian M. Scott Jan 8 '13 at 18:28
@BrianM.Scott: Excellent! – copper.hat Jan 8 '13 at 18:33

Simply substitute $ x=\pi/6 $ ( you have correctly discarded the other root) into

$$ 4 \sin x + \frac{9}{1+\sin x} $$

Use second derivative test to find out that it is minimum.

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