Question

Find the sub-interval(s) of \((0, \frac{\pi}{2})\) in which \(f(x) = \tan x - 4x\) is increasing.

Hint:

When dealing with \(\cos x\) in inequalities, remember its graph is decreasing in the first quadrant. A "smaller" value for cosine corresponds to a "larger" value for the angle.

Answer 1

Concept:
• A function \(f(x)\) is increasing in an interval if its derivative \(f'(x) > 0\) for all \(x\) in that interval.
• The derivative of \(\tan x\) is \(\sec^2 x\).

Step 1: Find the first derivative of the function
Given \(f(x) = \tan x - 4x\). Differentiating with respect to \(x\): \[ f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(4x) \] \[ f'(x) = \sec^2 x - 4 \]

Step 2: Set the condition for an increasing function
For \(f(x)\) to be increasing: \[ f'(x) > 0 \] \[ \sec^2 x - 4 > 0 \] \[ \sec^2 x > 4 \]

Step 3: Solve the inequality for \(x\)
Since \(\sec^2 x = \frac{1}{\cos^2 x}\): \[ \frac{1}{\cos^2 x} > 4 \] Taking reciprocal (reverses the inequality): \[ \cos^2 x < \frac{1}{4} \] In the interval \((0, \frac{\pi}{2})\), \(\cos x\) is positive. Taking the square root: \[ \cos x < \frac{1}{2} \]

Step 4: Determine the interval from the trigonometric condition
We know that \(\cos x = \frac{1}{2}\) at \(x = \frac{\pi}{3}\). As \(x\) increases from \(0\) to \(\frac{\pi}{2}\), the function \(\cos x\) decreases from \(1\) to \(0\). Therefore, \(\cos x\) will be less than \(\frac{1}{2}\) for values of \(x\) between \(\frac{\pi}{3}\) and \(\frac{\pi}{2}\). The required sub-interval is \((\frac{\pi}{3}, \frac{\pi}{2})\).