0 x maxx-2 x x+(1)/(3) 3x=

Question

$\underset{0\le x\le \pi }{max}{x-2\sin x\cos x+\frac{1}{3}\sin 3x}=$

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\le x\le \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}^{''}(\frac{5\pi }{6})=-2\sqrt{3}-\sqrt{3}
${f}\^{''}(\frac{\pi }{6})=2\sqrt{3}-\sqrt{3}>0$, 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:

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

Compare the values: The maximum value of $f(x)$ is given by $

Solutions · 1