3) $$ f(x)=2 x-4 \sin x $$ on $$ \left[0, \frac{\pi}{2}\right] $$

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UpStudy AI · Accepted Answer

**Step 1. Compute the derivative**

The derivative of 
$$
f(x)=2x-4\sin x
$$
is
$$
f'(x)=\frac{d}{dx}(2x)-\frac{d}{dx}(4\sin x)=2-4\cos x.
$$

**Step 2. Find the critical points**

Set the derivative equal to zero:
$$
2-4\cos x=0 \quad \Longrightarrow \quad \cos x=\frac{1}{2}.
$$
On the interval $$\left[0,\frac{\pi}{2}\right]$$ the solution to $$\cos x=\frac{1}{2}$$ is
$$
x=\frac{\pi}{3}.
$$

**Step 3. Evaluate $$f(x)$$ at the critical point and at the endpoints**

Calculate the function at the endpoints and the critical point.

- At $$x=0$$:
  $$
  f(0)=2\cdot 0-4\sin 0=0-0=0.
  $$

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

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

**Step 4. Determine which value is the minimum and which is the maximum**

Approximate the numerical values:
- $$f(0)=0$$.
- $$f\left(\frac{\pi}{3}\right)=\frac{2\pi}{3}-2\sqrt{3}\approx \frac{2\cdot 3.14}{3}-2\cdot1.73\approx 2.09-3.46\approx -1.37$$.
- $$f\left(\frac{\pi}{2}\right)=\pi-4\approx 3.14-4=-0.86$$.

Thus, the minimum value is approximately $$-1.37$$ at $$x=\frac{\pi}{3}$$, and the maximum value is $$0$$ at $$x=0$$.

**Final Answer:**

- Maximum: $$f(0)=0$$ at $$x=0$$.
- Minimum:  
  $$
  f\left(\frac{\pi}{3}\right)=\frac{2\pi}{3}-2\sqrt{3}
  $$
  at $$x=\frac{\pi}{3}$$.
The function $$ f(x) = 2x - 4\sin x $$ on the interval $$[0, \frac{\pi}{2}]$$ has: - **Maximum value** of $$0$$ at $$x = 0$$. - **Minimum value** of $$\frac{2\pi}{3} - 2\sqrt{3}$$ at $$x = \frac{\pi}{3}$$.

3) \( f(x)=2 x-4 \sin x \) on \( \left[0, \frac{\pi}{2}\right] \)

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