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I have a function, $f(x) = \sec x + \csc x$ on the interval $x \in (0, \pi/2)$.

The derivative of it is, $f'(x) = \csc^2(x) \sec^2(x) \left(\sin^3 x + \cos^3 x\right)$

Of course, when I tried to solve for $f'(x) = 0$, I realize that $f'(x)=0$ does not exist.

However, the question insists that the there is a minimum. Am I doing something wrong here?

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I don't mean to give you too much of the answer, but Wolfram Alpha is a really powerful tool when you need more insight about these kinds of problems: wolframalpha.com/input/?i=f%28x%29+%3D+sec+x+%2B+csc+x It appears your derivative is incorrect. –  Jonathan Rich Mar 14 '13 at 17:38
    
Hint: Derivative is: $tan(x) sec(x)-cot(x) csc(x)$, $x = \pi/4$ –  Amzoti Mar 14 '13 at 17:39
    
I expanded from tan(x)sec(x)−cot(x)csc(x) to (sin x)/(cos^2 x) + (cos x)/(sin^2 x). I am sorry for not formatting properly. –  0xFF Mar 14 '13 at 17:40
    
You lost a $-$ sign –  Robert Israel Mar 14 '13 at 17:42
    
(sin x)/(cos^2 x) - (cos x)/(sin^2 x) => (sin^3x-cos^3x)/(cos^2 x)(sin^2 x) => (sec^2)(csc^2 x)(sin^3x-cos^3x) Am I wrong with my expansion? –  0xFF Mar 14 '13 at 17:45
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3 Answers

up vote 3 down vote accepted

We should have $$\begin{align}f'(x) &= \frac{\sin x}{\cos^2 x}-\frac{\cos x}{\sin^2x}\\ &= \frac1{\cos^2x\sin^2x}\bigl(\sin^3x-\cos^3x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\bigl(\sin^2x+\sin x\cos x+\cos^2x\bigr)\\ &= \frac1{\cos^2x\sin^2x}(\sin x-\cos x)\left(1+\frac12\sin 2x\right).\end{align}$$ The fraction and the right-most factor at the end can't be zero for any real $x$. Hence, $f'(x)=0$ if and only if $\sin x=\cos x$. Where does that happen?

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The derivative is $0$ at $x=\frac{\pi}{4}$ as the derivative goes from $-$ to $+$ you really have a minimum.

The correct derivative is $$-\cot(x) \csc(x) + \sec(x) \tan(x)$$

enter image description here

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No roots of $f'(x)$ implies there is no local extrema. But on a closed interval, the minimum could also occur at a boundary point...

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That would be true. But in this case there is a zero of $f'(x)$ in the interval, and $f(x) \to +\infty$ as you approach the boundary points. –  Robert Israel Mar 14 '13 at 17:40
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