d the extrema of $$ f(x)=2 \sin x-\cos 2 x $$ the interval $$ [0,2 \pi] $$.

Question

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We start with the function
$$
f(x)=2\sin x-\cos 2x.
$$
Our goal is to find its extrema on the interval $$[0,2\pi]$$.

### Step 1. Find the derivative
Differentiate $$ f(x) $$ with respect to $$ x $$:
$$
f'(x)=\frac{d}{dx}\big(2\sin x\big)-\frac{d}{dx}\big(\cos 2x\big).
$$
The derivative of $$ 2\sin x $$ is $$ 2\cos x $$. For $$ \cos 2x $$, using the chain rule we have:
$$
\frac{d}{dx}\cos 2x = -\sin 2x \cdot 2 = -2\sin 2x.
$$
Thus,
$$
f'(x)=2\cos x - (-2\sin 2x)=2\cos x+2\sin 2x.
$$

### Step 2. Factor the derivative
Notice that $$\sin 2x$$ can be written as $$2\sin x\cos x$$. Therefore,
$$
f'(x)=2\cos x+2\cdot (2\sin x\cos x)=2\cos x+4\sin x\cos x.
$$
Factor out $$2\cos x$$:
$$
f'(x)=2\cos x\,(1+2\sin x).
$$

### Step 3. Solve for critical points
Set the derivative equal to zero:
$$
2\cos x\,(1+2\sin x)=0.
$$
This product is zero when either factor is zero.

1. **Case 1:** $$2\cos x=0$$  
$$
\cos x=0.
$$
In $$[0, 2\pi]$$, this occurs at
$$
x=\frac{\pi}{2}\quad \text{and} \quad x=\frac{3\pi}{2}.
$$

2. **Case 2:** $$1+2\sin x=0$$  
$$
2\sin x=-1 \quad\Longrightarrow\quad \sin x=-\frac{1}{2}.
$$
In $$[0, 2\pi]$$, $$\sin x=-\frac{1}{2}$$ when
$$
x=\frac{7\pi}{6}\quad \text{and} \quad x=\frac{11\pi}{6}.
$$

### Step 4. Evaluate $$ f(x) $$ at the critical points and endpoints
Include the endpoints $$x=0$$ and $$x=2\pi$$.

- **At $$ x=0 $$:**
  $$
  f(0)=2\sin 0-\cos (0)=0-1=-1.
  $$

- **At $$ x=\frac{\pi}{2} $$:**
  $$
  f\left(\frac{\pi}{2}\right)=2\sin\left(\frac{\pi}{2}\right)-\cos \left(2\cdot\frac{\pi}{2}\right)=2\cdot 1-\cos\pi=2-(-1)=3.
  $$

- **At $$ x=\frac{3\pi}{2} $$:**
  $$
  f\left(\frac{3\pi}{2}\right)=2\sin\left(\frac{3\pi}{2}\right)-\cos \left(2\cdot\frac{3\pi}{2}\right)=2\cdot(-1)-\cos (3\pi)=-2-(-1)=-1.
  $$
  (Since $$\cos 3\pi=\cos\pi=-1$$.)

- **At $$ x=\frac{7\pi}{6} $$:**  
  Recall that $$\sin\frac{7\pi}{6}=-\frac{1}{2}$$. Also,
  $$
  2\cdot\frac{7\pi}{6}=\frac{7\pi}{3}.
  $$
  Since $$\cos$$ is periodic with period $$2\pi$$, we have:
  $$
  \cos\left(\frac{7\pi}{3}\right)=\cos\left(\frac{7\pi}{3}-2\pi\right)=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}.
  $$
  Thus,
  $$
  f\left(\frac{7\pi}{6}\right)=2\left(-\frac{1}{2}\right)-\frac{1}{2}=-1-\frac{1}{2}=-\frac{3}{2}.
  $$

- **At $$ x=\frac{11\pi}{6} $$:**  
  Here, $$\sin\frac{11\pi}{6}=-\frac{1}{2}$$ and
  $$
  2\cdot\frac{11\pi}{6}=\frac{11\pi}{3}.
  $$
  Adjust by subtracting $$2\pi$$:
  $$
  \cos\left(\frac{11\pi}{3}\right)=\cos\left(\frac{11\pi}{3}-2\pi\right)=\cos\left(\frac{5\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}.
  $$
  Therefore,
  $$
  f\left(\frac{11\pi}{6}\right)=2\left(-\frac{1}{2}\right)-\frac{1}{2}=-1-\frac{1}{2}=-\frac{3}{2}.
  $$

- **At $$ x=2\pi $$:**
  $$
  f(2\pi)=2\sin(2\pi)-\cos (4\pi)=0-1=-1.
  $$

### Step 5. Identify the extrema
- The maximum value is
  $$
  f\left(\frac{\pi}{2}\right)=3,
  $$
  so the absolute maximum is $$ 3 $$ at $$ x=\frac{\pi}{2} $$.

- The minimum value is
  $$
  f\left(\frac{7\pi}{6}\right)=f\left(\frac{11\pi}{6}\right)=-\frac{3}{2},
  $$
  so the absolute minimum is $$ -\frac{3}{2} $$ at $$ x=\frac{7\pi}{6} $$ and $$ x=\frac{11\pi}{6} $$.

### Final Answer
- Absolute maximum: $$ f\left(\frac{\pi}{2}\right)=3 $$.
- Absolute minimum: $$ f\left(\frac{7\pi}{6}\right)=f\left(\frac{11\pi}{6}\right)=-\frac{3}{2} $$.
The function $$ f(x) = 2\sin x - \cos 2x $$ has an absolute maximum of 3 at $$ x = \frac{\pi}{2} $$ and an absolute minimum of $$ -\frac{3}{2} $$ at $$ x = \frac{7\pi}{6} $$ and $$ x = \frac{11\pi}{6} $$ within the interval $$[0, 2\pi]$$.

d the extrema of

the interval .

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Solución

Step 1. Find the derivative

Step 2. Factor the derivative

Step 3. Solve for critical points

Step 4. Evaluate at the critical points and endpoints

Step 5. Identify the extrema

Final Answer

Beyond the Answer

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