What Are the Critical Numbers for the Function f(θ) = 2sec(θ) + tan(θ)?

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Homework Statement



f(theta)=2sec(theta)+tan(theta)

The Attempt at a Solution


I found the derivative and set it equal to zero and to reduce writing I substituted x for theta
f'(x)=2sec(x)tan(x)+sec^2(x)
sec(x)[2tan(x)+sec(x)]=0

My question is what are the critical numbers? do the critical numbers exist where sec(x) is underfined because sec(x) will never equal 0.
 
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The expression can be further simplified:
<br /> \sec^{2}x(2\sin x+1)=0<br />
as sec(x) is never zero, we can divide through by it and not loose any solutions. What are you left with?
 
sec(x)[2tan(x)+sec(x)] = 0

You are trying to find an x to make the left side equal to 0.

Like you said, there is no x for which sec(x) = 0, so you can discard that possibility.

How else can the left side of that equation be 0?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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