First of all finding the critical numbers is a problem for this one: $\displaystyle f(x) = 3\sin(x) + 3\cos(x)$ $\displaystyle f'(x) = 3\cos(x) + [-3\sin(x)]$ $\displaystyle f'(x) = 3\cos(x) - 3\sin(x)$ Next move, hint?
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Originally Posted by Jason76 First of all finding the critical numbers is a problem for this one: $\displaystyle f(x) = 3\sin(x) + 3\cos(x)$ $\displaystyle f'(x) = 3\cos(x) + [-3\sin(x)]$ $\displaystyle f'(x) = 3\cos(x) - 3\sin(x)$ Next move, hint? Hello, I assume that you want to solve for x $\displaystyle 3\cos(x) - 3\sin(x) = 0$ If so: $\displaystyle 3\cos(x) - 3\sin(x) = 0~\implies~1 = \frac{3 \sin(x)}{3 \cos(x)} = \tan(x)$ and therefore: $\displaystyle x = \tan^{-1}(1)$
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