Question

Find the critical points of the function \(f(x)=x-\ell n x+\int_{2}^{x}\left(\frac{1}{z}-2-2 \cos 4 z\right) d z\)

Short answer

The critical points of the function are given by the following expressions: $$ x = \frac{\cos^{-1}\left(-\frac{1}{2}\right)}{4} + \frac{n\pi}{2} \quad \text{or} \quad x = -\frac{\cos^{-1}\left(-\frac{1}{2}\right)}{4} + \frac{(2n + 1)\pi}{4},$$ where \(n\) is an integer.

Step-by-step solution

Differentiate the function

Differentiate the given function \(f(x)=x-\ell n x+\int_{2}^{x}\left(\frac{1}{z}-2-2 \cos 4 z\right) d z\) with respect to \(x\): $$ f'(x) = \frac{d}{dx} \left( x - \ell n x + \int_{2}^{x} \left( \frac{1}{z} - 2 - 2 \cos 4z \right) dz \right) $$

Apply differentiation rules

We apply the sum rule, constant rule, and chain rule to differentiate each term individually: $$ f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\ell n x) + \frac{d}{dx} \left( \int_{2}^{x} \left( \frac{1}{z} - 2 - 2 \cos 4z \right) dz \right) $$

Differentiate each term

Differentiate \(x\), \(\ell n x\), and the integral term: $$ f'(x) = 1 - \frac{1}{x} + \left( \frac{1}{x} - 2 - 2 \cos 4x \right) $$

Simplify \(f'(x)\)

Simplify the expression for the first derivative: $$ f'(x) = 1 - 2 - 2 \cos 4x $$

Find the critical points

Set the first derivative equal to 0 to find the critical points: $$ 0 = 1 - 2 - 2 \cos 4x $$

Solve for x

Solve for x to find the critical points: $$ 2 \cos 4x = 1 - 2 $$ $$ \cos 4x = -\frac{1}{2} $$ To find the values of \(x\) that satisfy this equation, we can use the inverse cosine function: $$ 4x = \cos^{-1}\left(-\frac{1}{2}\right) + 2n\pi \quad \text{or} \quad 4x = -\cos^{-1}\left(-\frac{1}{2}\right) +(2n+1)\pi $$ Divide both sides by 4: $$ x = \frac{\cos^{-1}\left(-\frac{1}{2}\right)}{4} + \frac{n\pi}{2} \quad \text{or} \quad x = -\frac{\cos^{-1}\left(-\frac{1}{2}\right)}{4} + \frac{(2n + 1)\pi}{4} $$ These expressions give us the critical points of the function.