[ _0 x x-2 x x+(1)/(3) 3 x=]

Question

$$\max _\limits{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$$

Accepted Answer

Given the function:

$$
f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x}
$$

We want to find the maximum value of this expression for $$0 \leq x \leq \pi$$.

Rewrite the expressio: 1: Rewrite the expression

Notice that we can rewrite the expression as:

$$
f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}
$$

Find the first derivativ: 2: Find the first derivative

Now, let's find the derivative of this expression with respect to $x$:

$$
f'(x) = 1 - 2\cos{2x} + \cos{3x}
$$

Find the critical point: 3: Find the critical points

To find the critical points, we set $f'(x) = 0$:

$$
1 - 2\cos{2x} + \cos{3x} = 0
$$

Solve for : 4: Solve for x

This equation can be rewritten as:

$$
(2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0
$$

We get three possible solutions for $x$:

$$
\cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1
$$

Which gives us:
$$
x = \frac{5\pi}{6}, \frac{\pi}{6}, 0
$$

Find the second derivativ: 5: Find the second derivative

Now, let's find the second derivative of the expression:

$$
f''(x) = 4\sin{2x} - 3\sin{3x}
$$

Analyze the critical points: Evaluate the second derivative at the critical points:

$$f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0$$, indicating a local maximum.
$$f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$$, indicating a local minimum.
$f''(0) = 0$, inconclusive.

Evaluate the function at the local maximum point and the boundary points: Since we are looking for the maximum value of $f(x)$, we can now evaluate the function at the local maximum point and the boundary points:

$$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$
$f(0) = 0$
$$f(\pi) = \pi$$

Compare the values: The maximum value of $f(x)$ is given by $$f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$.

Solutions · 1