# What are the critical values, if any, of  f(x)=cscxtanx-sqrt(xcosx) in [0,2pi]?

May 22, 2017

$x \in \left\{0 , 0.455 , 2 \pi\right\}$

#### Explanation:

Critical values are points where $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ is undefined, BUT $f \left(x\right)$ is defined.
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First of all, $\csc x \tan x = \frac{1}{\sin} x \cdot \sin \frac{x}{\cos} x = \frac{1}{\cos} x = \sec x$, so we can substitute $\sec x$ into the equation.

$f \left(x\right) = \sec x - \sqrt{x \cos x}$

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First, we must find points where $f ' \left(x\right) = 0$

$f ' \left(x\right) = \sec x \tan x - \frac{\cos x - x \sin x}{2 \sqrt{x \cos x}} = 0$

Graphing this function shows that the only solution on the interval $\left[0 , 2 \pi\right]$ is $0.455$.

Next, we find points where $f ' \left(x\right)$ is undefined. By using a calculator, or by manipulating the equation, we can see that $f ' \left(x\right)$ is not defined on the ENTIRE interval $\left[\frac{\pi}{2} , \frac{3 \pi}{2}\right]$. This would normally mean that $x = \frac{\pi}{2}$ and $x = \frac{3 \pi}{2}$ are critical points, but $f \left(\frac{\pi}{2}\right)$and$f \left(\frac{3 \pi}{2}\right)$are also not defined, since $\sec x$ has asymptotes at those points.

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Therefore, our critical points are the endpoints $x = 0$ and $x = 2 \pi$ as well as the point $x = 0.455$.