Question

Find the interval in which the function \(f(x)=\dfrac{4\sin x-2x-x\cos x}{2+\cos x}\) is strictly increasing and strictly decreasing.

Hint:

Differentiate f(x) with the quotient rule, simplify using sin squared x plus cos squared x equals 1, then check the sign of cos x.

Answer 1

Step 1: Understanding the Concept:
A function is strictly increasing where its derivative is positive and strictly decreasing where its derivative is negative.
Since f is built from sin x and cos x, which repeat every \(2\pi\), it is enough to study one full period, so we take \(x\in[0,2\pi]\).

Step 2: Differentiate Using the Quotient Rule:
Let \(u=4\sin x-2x-x\cos x\) and \(v=2+\cos x\), so \(f=u/v\).
Differentiate u using the product rule on the term \(x\cos x\), and differentiate v directly.
\[ u'=4\cos x-2-(\cos x-x\sin x)=3\cos x-2+x\sin x, \qquad v'=-\sin x \]

Step 3: Simplify f'(x):
Apply the quotient rule \(f'=\dfrac{u'v-uv'}{v^2}\) and expand the numerator.
The numerator works out to \(4\sin^2x+3\cos^2x+4\cos x-4\), which simplifies using \(\sin^2x=1-\cos^2x\).
\[ 4(1-\cos^2x)+3\cos^2x+4\cos x-4=-\cos^2x+4\cos x=\cos x(4-\cos x) \]
\[ f'(x)=\frac{\cos x(4-\cos x)}{(2+\cos x)^2} \]

Step 4: Study the Sign of f'(x):
Since \(\cos x\) always lies between \(-1\) and \(1\), the factor \(4-\cos x\) is always positive, and \((2+\cos x)^2\) is always positive.
So the sign of \(f'(x)\) is exactly the sign of \(\cos x\), nothing else affects it.
On \([0,2\pi]\), \(\cos x>0\) on \((0,\pi/2)\cup(3\pi/2,2\pi)\) and \(\cos x<0\) on \((\pi/2,3\pi/2)\).

Final Answer:
f is strictly increasing where cos x is positive and strictly decreasing where cos x is negative.
\[ \boxed{\text{Increasing on } \left(0,\tfrac{\pi}{2}\right)\cup\left(\tfrac{3\pi}{2},2\pi\right), \quad \text{Decreasing on } \left(\tfrac{\pi}{2},\tfrac{3\pi}{2}\right)} \]