Question

The function \(f(x) = 4 \sin^3 x - 6 \sin^2 x + 12 \sin x + 100\) is strictly

• A. decreasing in \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\)
• B. increasing in \([ \pi, \frac{3\pi}{2} ]\)
• C. decreasing in \([0, \frac{\pi}{2}]\)
• D. decreasing in \([\frac{\pi}{2}, \pi]\)

Answer 1

The Correct Option is D

Given the function:

\[ f(x) = 4 \sin^3(x) - 6 \sin^2(x) + 12 \sin(x) + 100 \]

To find the derivative, we use the chain rule and the power rule:

\[ f'(x) = 3(4 \sin^2(x) \cos(x)) - 2(6 \sin(x) \cos(x)) + 12 \cos(x) \]

Simplifying further:

\[ f'(x) = 12 \sin^2(x) \cos(x) - 12 \sin(x) \cos(x) + 12 \cos(x) \]

Now, to determine the intervals of increasing or decreasing, we need to analyze the sign of \(f'(x)\).

Let's analyze the sign of \(f'(x)\) in different intervals:

In the interval \(\left[\frac{\pi}{2}, \pi\right]\), for \(x\) values between \(\frac{\pi}{2}\) and \(\pi\), \(\sin(x)\) is positive, and \(\cos(x)\) is negative.

Since \(\sin^2(x)\) and \(\cos(x)\) are non-negative, while \(\sin(x)\) is positive, all terms in \(f'(x)\) are positive.

Therefore, \(f'(x)\) is always positive in this interval. Thus, \(f(x)\) is strictly increasing in the interval \(\left[\frac{\pi}{2}, \pi\right]\).

Based on the analysis, we can conclude that the function

\[ f(x) = 4\sin^3(x) - 6\sin^2(x) + 12\sin(x) + 100 \]

is strictly increasing in the interval \(\left[\frac{\pi}{2}, \pi\right]\), which corresponds to option (D) decreasing in \(\left[\frac{\pi}{2}, \pi\right]\).

Answer 2

Given the function \(f(x) = 4 \sin^3 x - 6 \sin^2 x + 12 \sin x + 100\), we want to find where it is strictly increasing or decreasing.

First, find the derivative of f(x) with respect to x:

\(f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x\)

\(f'(x) = 12 \cos x (\sin^2 x - \sin x + 1)\)

Now, we need to determine the sign of f'(x) in the given intervals.

Let \(g(x) = \sin^2 x - \sin x + 1\). We can rewrite this as a quadratic in sin x:

\(g(x) = (\sin x)^2 - (\sin x) + 1\)

To find the minimum value of this quadratic, we can complete the square or use the vertex formula:

g(x) = \((\sin x - \frac{1}{2})^2 + \frac{3}{4}\)

Since \((\sin x - \frac{1}{2})^2\) is always non-negative, and \(\frac{3}{4}\) is positive, g(x) is always positive. Thus, \(g(x) > 0\) for all x.

Therefore, the sign of f'(x) depends only on the sign of cos x.

• In \(\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\), \(\cos x \geq 0\). Thus, \(f'(x) \geq 0\) in this interval. The function is increasing except at \(x= \pm \frac{\pi}{2}\) so not strictly decreasing.
• In \([ \pi, \frac{3\pi}{2} ]\), \(\cos x \leq 0\). Thus, \(f'(x) \leq 0\) in this interval, however there are parts where \(f'(x) = 0\) so not strictly increasing.
• In \([0, \frac{\pi}{2}]\), \(\cos x \geq 0\). Thus, \(f'(x) \geq 0\) in this interval so it must be increasing and not decreasing.
• In \([\frac{\pi}{2}, \pi]\), \(\cos x \leq 0\). Thus, \(f'(x) \leq 0\) in this interval. The function is decreasing except at \(x=\frac{\pi}{2}\) so not strictly decreasing

Since g(x) > 0, then when \(cos(x) >=0\), f'(x) is >0, meaning it is strictly increasing.

Answer: increasing in \(\left[\frac{\pi}{2}, \pi\right]\)