To solve this problem, let's break down the steps required to find the local and absolute maxima and minima for the function:

$f(x) = 8x - 16 \cos(x)$ on the interval $[-π, π]$.

Step 1: Find the Derivative of $f(x)$

To identify critical points, we need the first derivative of $f(x)$: $f'(x) = \frac{d}{dx}(8x - 16 \cos(x)) = 8 + 16 \sin(x)$

Step 2: Set $f'(x) = 0$ and Solve for $x$

To find critical points, set $f'(x) = 0$: $8 + 16 \sin(x) = 0$ $16 \sin(x) = -8$ $\sin(x) = -\frac{1}{2}$

Within the interval $[-π, π]$, $\sin(x) = -\frac{1}{2}$ at: x = -\frac{\pi}{6} \quad \text{and} \quad x = \frac{7\pi}{6} \quad \text{(or, equivalently, x = \frac{\pi}{6} - π = -\frac{5\pi}{6})}

Thus, our critical points are $x = -\frac{\pi}{6}$ and $x = \frac{\pi}{6}$.

Step 3: Evaluate $f(x)$ at Critical Points and Endpoints

To determine the nature of these critical points and check for absolute maxima and minima, evaluate $f(x)$ at the critical points and endpoints $x = -π$ and $x = π$.

1. At $x = -π$: $f(-π) = 8(-π) - 16 \cos(-π) = -8π - 16(-1) = -8π + 16$

2. At $x = π$: $f(π) = 8(π) - 16 \cos(π) = 8π - 16(-1) = 8π + 16$

3. At $x = -\frac{\pi}{6}$: $f\left(-\frac{\pi}{6}\right) = 8\left(-\frac{\pi}{6}\right) - 16 \cos\left(-\frac{\pi}{6}\right) = -\frac{4\pi}{3} - 16 \cdot \frac{\sqrt{3}}{2} = -\frac{4\pi}{3} - 8\sqrt{3}$

4. At $x = \frac{\pi}{6}$: $f\left(\frac{\pi}{6}\right) = 8\left(\frac{\pi}{6}\right) - 16 \cos\left(\frac{\pi}{6}\right) = \frac{4\pi}{3} - 8\sqrt{3}$

Step 4: Compare Values to Determine Maxima and Minima

Now we compare all computed values:

• $f(-π) = -8π + 16$
• $f(π) = 8π + 16$
• $f\left(-\frac{\pi}{6}\right) = -\frac{4\pi}{3} - 8\sqrt{3}$
• $f\left(\frac{\pi}{6}\right) = \frac{4\pi}{3} - 8\sqrt{3}$

From these calculations:

• The absolute maximum occurs at $x = π$ with value $f(π) = 8π + 16$.
• The absolute minimum occurs at $x = -π$ with value $f(-π) = -8π + 16$.
• There are no other local maxima or minima within the interval $[-π, π]$, since the values at $x = -\frac{\pi}{6}$ and $x = \frac{\pi}{6}$ are neither the highest nor the lowest.

Final Answer

• Local maxima: DNE
• Local minima: DNE
• Absolute maxima: $x = π$
• Absolute minima: $x = -π$

Would you like a more detailed explanation or have any questions?

Related Questions

1. How do we determine if a critical point is a local extremum?
2. Why do we need to evaluate the function at the endpoints in addition to the critical points?
3. What are the implications if $f(x)$ had no critical points in the interval?
4. How can the second derivative test help in finding maxima and minima?
5. What role does periodicity play in trigonometric functions when finding extrema?

Tip

Always evaluate the endpoints of the interval in closed-interval problems, as extrema may occur there.