Determination of Extrema

Solution Approach: Use the derivative method to find extrema, combined with graphical method for verification.

Detailed Steps:

1. Find the derivative: $f'(x) = 3x^2 - 6x = 3x(x - 2)$

2. Find stationary points: Set $f'(x) = 0$, get $x = 0$ or $x = 2$

3. Determine extrema type:

   • When $x < 0$, $f'(x) > 0$ (function increasing)
   • When $0 < x < 2$, $f'(x) < 0$ (function decreasing)
   • When $x > 2$, $f'(x) > 0$ (function increasing)

4. Conclusion:

   • $x = 0$ is a maximum point
   • $x = 2$ is a minimum point

5. Calculate extrema values:

   • $f(0) = 2$ (maximum)
   • $f(2) = 8 - 12 + 2 = -2$ (minimum)

Answer: The function has a maximum value of $2$ at $x = 0$ and a minimum value of $-2$ at $x = 2$.

Solution Approach: This is a non-differentiable function that requires special handling.

Detailed Steps:

1. Analyze function properties: $f(x) = |x|$ is non-differentiable at $x = 0$ but continuous.

2. Use numerical comparison method:

   • $f(0) = 0$
   • For any $x \neq 0$, $f(x) > 0$

3. Conclusion: $x = 0$ is a minimum point with minimum value $0$.

4. Verification: The function reaches its global minimum at $x = 0$.

Answer: The function has a minimum value of $0$ at $x = 0$.

Solution Approach: Use the derivative method to find extrema.

Detailed Steps:

1. Find the derivative: $f'(x) = 4x^3 - 8x = 4x(x^2 - 2)$

2. Find stationary points: Set $f'(x) = 0$, get $x = 0$ or $x = \pm\sqrt{2}$

3. Determine extrema type:

   • When $x < -\sqrt{2}$, $f'(x) < 0$
   • When $-\sqrt{2} < x < 0$, $f'(x) > 0$
   • When $0 < x < \sqrt{2}$, $f'(x) < 0$
   • When $x > \sqrt{2}$, $f'(x) > 0$

4. Conclusion:

   • $x = -\sqrt{2}$ is a minimum point
   • $x = 0$ is a maximum point
   • $x = \sqrt{2}$ is a minimum point

5. Calculate extrema values:

   • $f(-\sqrt{2}) = 4 - 8 = -4$ (minimum)
   • $f(0) = 0$ (maximum)
   • $f(\sqrt{2}) = 4 - 8 = -4$ (minimum)

Answer: The function has a maximum value of $0$ at $x = 0$ and minimum values of $-4$ at $x = \pm\sqrt{2}$.

Solution Approach: Use the derivative method to find extrema.

Detailed Steps:

1. Find the derivative: $f'(x) = \cos x - \sin x$

2. Find stationary points: Set $f'(x) = 0$, get $\cos x = \sin x$ That is $\tan x = 1$, so $x = \frac{\pi}{4} + k\pi$, $k \in \mathbb{Z}$ On $[0, 2\pi]$, stationary points are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$

3. Determine extrema type:

   • When $0 < x < \frac{\pi}{4}$, $f'(x) > 0$
   • When $\frac{\pi}{4} < x < \frac{5\pi}{4}$, $f'(x) < 0$
   • When $\frac{5\pi}{4} < x < 2\pi$, $f'(x) > 0$

4. Conclusion:

   • $x = \frac{\pi}{4}$ is a maximum point
   • $x = \frac{5\pi}{4}$ is a minimum point

5. Calculate extrema values:

   • $f(\frac{\pi}{4}) = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$ (maximum)
   • $f(\frac{5\pi}{4}) = \sin\frac{5\pi}{4} + \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$ (minimum)

Answer: The function has a maximum value of $\sqrt{2}$ at $x = \frac{\pi}{4}$ and a minimum value of $-\sqrt{2}$ at $x = \frac{5\pi}{4}$.