Question

Consider the function f(x) defined as:

(b) At what values of x does the derivative of f(x) equal zero? Justify your answer.

(c) Determine the intervals on which the function f(x) is increasing or decreasing. Justify your answer.

Answer

(a) To find the derivative of f(x) with respect to x, we can use the product rule and the chain rule.

Let's break down the function f(x) into its individual parts:

Now, applying the product rule:

Differentiating each term:

Finally, simplifying the expression:

Therefore, the derivative of f(x) with respect to x is:

This equation involves a product of two factors, so it can be satisfied if either of the factors is zero.

Setting the first factor $9x^2 \cdot \sin(2x)$ equal to zero:

The equation $9x^2 = 0$ has only one solution: $x = 0$.

The equation $\sin(2x) = 0$ has infinitely many solutions. We know that $\sin(2x) = 0$ when $2x = n\pi$, where n is any integer. So, $x = \frac{n\pi}{2}$ for all integers n.

Setting the second factor $6x^3 \cdot \cos(2x)$ equal to zero:

The equation $6x^3 = 0$ has only one solution: $x = 0$.

The equation $\cos(2x) = 0$ has infinitely many solutions. We know that $\cos(2x) = 0$ when $2x = \frac{\pi}{2} + n\pi$, where n is any integer. So, $x = \frac{\pi}{4} + \frac{n\pi}{2}$ for all integers n.

Therefore, the values of x at which the derivative of f(x) equals zero are: x = 0, x = π/4, x = π/2, x = 5π/4, x = 3π/2, x = 9π/4, ...

(c) To determine the intervals on which the function f(x) is increasing or decreasing, we can analyze the sign of the derivative f'(x) in different intervals.

• For x < 0, both $9x^2 \cdot \sin(2x)$ and $6x^3 \cdot \cos(2x)$ are negative. So, the function f(x) is decreasing on this interval.

• For 0 < x < π/4, $9x^2 \cdot \sin(2x)$ is positive and $6x^3 \cdot \cos(2x)$ is negative. So, the function f(x) is increasing on this interval.

• For π/4 < x < π/2, both $9x^2 \cdot \sin(2x)$ and $6x^3 \cdot \cos(2x)$ are positive. So, the function f(x) is increasing on this interval.

• For π/2 < x < 5π/4, $9x^2 \cdot \sin(2x)$ is positive and $6x^3 \cdot \cos(2x)$ is negative. So, the function f(x) is increasing on this interval.

• For 5π/4 < x < π, both $9x^2 \cdot \sin(2x)$ and $6x^3 \cdot \cos(2x)$ are negative. So, the function f(x) is decreasing on this interval.

• Similar patterns repeat for intervals between successive solutions x = π/2, x = π, x = 5π/4, and so on.

Therefore, the function f(x) is increasing on the intervals (0, π/4) and (π/2, 5π/4), and it is decreasing on the intervals (π/4, π/2) and (5π/4, π).