Key Concepts and Formulas

• First Derivative Test for Increasing/Decreasing Functions: A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ on that interval, and decreasing if $f'(x) \le 0$ on that interval.
• Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
• Trigonometric Derivatives: $\frac{d}{dx}(\sin x) = \cos x$

Step-by-Step Solution

Step 1: Find the derivative of f(x)

We need to find $f'(x)$ to determine the intervals of monotonicity. $f(x) = 3\sin^4 x + 10\sin^3 x + 6\sin^2 x - 3$ $f'(x) = \frac{d}{dx}(3\sin^4 x + 10\sin^3 x + 6\sin^2 x - 3)$ Using the chain rule, we get: $f'(x) = 3(4\sin^3 x)(\cos x) + 10(3\sin^2 x)(\cos x) + 6(2\sin x)(\cos x) - 0$ $f'(x) = 12\sin^3 x \cos x + 30\sin^2 x \cos x + 12\sin x \cos x$

Step 2: Factor the derivative

Factoring out common terms simplifies the expression and helps us find critical points. $f'(x) = 6\sin x \cos x (2\sin^2 x + 5\sin x + 2)$ $f'(x) = 6\sin x \cos x (2\sin x + 1)(\sin x + 2)$

Step 3: Analyze the sign of f'(x) in the given interval

We are given the interval $x \in \left[ -\frac{\pi}{6}, \frac{\pi}{2} \right]$. We need to determine where $f'(x) \ge 0$ or $f'(x) \le 0$ in this interval.

• $6 > 0$
• Since $x \in \left[ -\frac{\pi}{6}, \frac{\pi}{2} \right]$, $\cos x \ge 0$.
• $(\sin x + 2)$ is always positive since $\sin x$ ranges from -1 to 1.
• We now analyze the sign of $\sin x$ and $(2\sin x + 1)$.

In the interval $\left( -\frac{\pi}{6}, 0 \right)$, $\sin x < 0$ and $2\sin x + 1 > 0$ because $\sin x > -\frac{1}{2}$. So, $f'(x) = 6\sin x \cos x (2\sin x + 1)(\sin x + 2) < 0$.

In the interval $\left( 0, \frac{\pi}{2} \right)$, $\sin x > 0$, so $f'(x) = 6\sin x \cos x (2\sin x + 1)(\sin x + 2) > 0$.

At $x = -\frac{\pi}{6}$, $\sin x = -\frac{1}{2}$, so $2\sin x + 1 = 0$, and thus $f'(-\frac{\pi}{6}) = 0$. At $x = 0$, $\sin x = 0$, so $f'(0) = 0$. At $x = \frac{\pi}{2}$, $\cos x = 0$, so $f'(\frac{\pi}{2}) = 0$.

Step 4: Determine the intervals of increasing and decreasing behavior.

Based on the sign analysis of $f'(x)$:

• In $\left( -\frac{\pi}{6}, 0 \right)$, $f'(x) < 0$, so $f(x)$ is decreasing.
• In $\left( 0, \frac{\pi}{2} \right)$, $f'(x) > 0$, so $f(x)$ is increasing.

Since $f'(x) \ge 0$ on $\left[ -\frac{\pi}{6}, \frac{\pi}{2} \right]$ (except at $x = -\frac{\pi}{6}, 0, \frac{\pi}{2}$), the function is increasing on $\left( -\frac{\pi}{6}, \frac{\pi}{2} \right)$. Note that $f'(x)$ is not identically zero on any subinterval of $\left( -\frac{\pi}{6}, \frac{\pi}{2} \right)$.

Common Mistakes & Tips

• Remember to use the chain rule when differentiating composite functions like $\sin^n x$.
• Factoring $f'(x)$ makes it easier to analyze its sign.
• Don't forget to consider the given interval when determining the sign of $f'(x)$.

Summary

We found the derivative of the given function, factored it, and analyzed its sign within the given interval. This analysis showed that the function is increasing on the interval $\left( -\frac{\pi}{6}, \frac{\pi}{2} \right)$.

Final Answer

The final answer is \boxed{increasing in $\left( { - {\pi \over 6},{\pi \over 2}} \right)$} which corresponds to option (A).