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

1 Answer
John D.
May 22, 2017

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)andf((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.

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