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

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We wish to study the function
$$
f(x)=2x-4\sin x
$$
on the interval $$\left[0,\frac{\pi}{2}\right]$$. A common task is to find the extrema of $$f$$ on this interval.

**Step 1. Compute the derivative**

Differentiate $$f(x)$$ with respect to $$x$$:
$$
f'(x)=\frac{d}{dx}(2x)-4\frac{d}{dx}(\sin x)=2-4\cos x.
$$

**Step 2. Solve for the critical points**

Critical points occur when $$f'(x)=0$$ or when $$f'(x)$$ does not exist. Here, $$f'(x)$$ exists everywhere. Set
$$
2-4\cos x=0.
$$
Solve for $$\cos x$$:
$$
4\cos x=2\quad\Longrightarrow\quad\cos x=\frac{1}{2}.
$$
Within 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 the endpoints**

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

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

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

**Step 4. Compare the values to determine the extrema**

Let us compare the three values:

- $$f(0)=0$$.
- $$f\left(\frac{\pi}{3}\right)=\frac{2\pi}{3}-2\sqrt{3}$$.  
  In approximate numerical values:
  $$
  \frac{2\pi}{3}\approx2.094\quad\text{and}\quad2\sqrt{3}\approx3.464,
  $$
  so
  $$
  f\left(\frac{\pi}{3}\right)\approx2.094-3.464\approx-1.37.
  $$
- $$f\left(\frac{\pi}{2}\right)=\pi-4$$.  
  In approximate value:
  $$
  \pi\approx3.142,\quad \pi-4\approx -0.858.
  $$

Thus, we have:
- Maximum value: $$f(0)=0$$.
- Minimum value: $$f\left(\frac{\pi}{3}\right)=\frac{2\pi}{3}-2\sqrt{3}$$.

**Final Answer**

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

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

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