EVT Find the maximum and minimum values of $$ f(x)=x=2 \sin x $$ on the interval $$ [0,2 \pi] $$

Question

UpStudy AI · Accepted Answer

Let
$$
f(x)=x+2\sin x
$$
on the interval $$[0,2\pi]$$.

**Step 1. Find the derivative**

We compute
$$
f'(x)=\frac{d}{dx}\left(x+2\sin x\right)=1+2\cos x.
$$

**Step 2. Find critical points**

Critical points occur when $$f'(x)=0$$ or when $$f'(x)$$ is undefined. Since $$f'(x)=1+2\cos x$$ is defined everywhere, we need to solve:
$$
1+2\cos x=0 \quad\Longrightarrow\quad \cos x = -\frac{1}{2}.
$$
Within the interval $$[0,2\pi]$$, the solutions of $$\cos x = -\frac{1}{2}$$ are:
$$
x=\frac{2\pi}{3}\quad \text{and}\quad x=\frac{4\pi}{3}.
$$

**Step 3. Evaluate the function at the critical points and endpoints**

- At $$x=0$$:
  $$
  f(0)=0+2\sin 0=0.
  $$
- At $$x=\frac{2\pi}{3}$$:
  $$
  f\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}+2\sin\frac{2\pi}{3}=\frac{2\pi}{3}+2\cdot\frac{\sqrt{3}}{2}=\frac{2\pi}{3}+\sqrt{3}.
  $$
- At $$x=\frac{4\pi}{3}$$:
  $$
  f\left(\frac{4\pi}{3}\right)=\frac{4\pi}{3}+2\sin\frac{4\pi}{3}=\frac{4\pi}{3}+2\cdot\left(-\frac{\sqrt{3}}{2}\right)=\frac{4\pi}{3}-\sqrt{3}.
  $$
- At $$x=2\pi$$:
  $$
  f(2\pi)=2\pi+2\sin2\pi=2\pi.
  $$

**Step 4. Compare the values**

Let us approximate the numerical values to compare:
- $$f(0)=0$$.
- $$f\left(\frac{2\pi}{3}\right) \approx \frac{2\pi}{3}+\sqrt{3}\approx 2.094+1.732\approx 3.826$$.
- $$f\left(\frac{4\pi}{3}\right) \approx \frac{4\pi}{3}-\sqrt{3}\approx 4.189-1.732\approx 2.457$$.
- $$f(2\pi)\approx 6.283$$.

Thus, the smallest value is $$f(0)=0$$ and the largest value is $$f(2\pi)=2\pi$$.

**Final Answer**

The minimum value of $$f(x)$$ on $$[0,2\pi]$$ is $$0$$ (at $$x=0$$), and the maximum value is $$2\pi$$ (at $$x=2\pi$$).
The minimum value of $$ f(x) = x + 2\sin x $$ on the interval $$[0, 2\pi]$$ is $$0$$ at $$x=0$$, and the maximum value is $$2\pi$$ at $$x=2\pi$$.

EVT Find the maximum and minimum values of on the interval

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