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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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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 $$-\cot(x) \csc(x) + \sec(x) \tan(x)$$
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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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