Question

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

Answer 1

`x in {0, 0.455, 2pi}`

Explanation:

Critical values are points where `f'(x)=0` or `f'(x)` is undefined, BUT `f(x)` is defined.
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First of all, `cscxtanx=1/sinx*sinx/cosx=1/cosx=secx`, so we can substitute `secx` into the equation.

`f(x)=secx-sqrt(xcosx)`

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First, we must find points where `f'(x)=0`

`f'(x) = secxtanx-(cosx-xsinx)/(2sqrt(xcosx))=0`

Graphing this function shows that the only solution on the interval `[0,2pi]` is `0.455`.

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

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