To solve this problem and find the absolute extreme values of $f(x) = (x+2)^{\frac{4}{3}}$ on the interval $[-6, 6]$, we can use the following steps:

Step 1: Evaluate the function at the endpoints

The endpoints of the interval are $x = -6$ and $x = 6$. Let's calculate $f(-6)$ and $f(6)$.

1. At $x = -6$: $f(-6) = (-6 + 2)^{\frac{4}{3}} = (-4)^{\frac{4}{3}}$ Calculating $(-4)^{\frac{4}{3}}$, we raise $-4$ to the power of $\frac{4}{3}$: $(-4)^{\frac{4}{3}} = \left(4^{\frac{4}{3}}\right) = 16$ So, $f(-6) = 16$.

2. At $x = 6$: $f(6) = (6 + 2)^{\frac{4}{3}} = 8^{\frac{4}{3}}$ Similarly, $8^{\frac{4}{3}} = 16$, so $f(6) = 16$.

Step 2: Find Critical Points by Setting the Derivative to Zero

To find critical points within the interval, we need to compute the derivative of $f(x)$ and set it to zero.

The derivative of $f(x) = (x+2)^{\frac{4}{3}}$ is: $f'(x) = \frac{4}{3}(x+2)^{\frac{1}{3}}$ Now, set $f'(x) = 0$: $\frac{4}{3}(x+2)^{\frac{1}{3}} = 0$ This equation is zero when $(x+2)^{\frac{1}{3}} = 0$, which implies $x = -2$.

Step 3: Evaluate $f(x)$ at the Critical Point

Now, evaluate $f(x)$ at $x = -2$: $f(-2) = (-2 + 2)^{\frac{4}{3}} = 0^{\frac{4}{3}} = 0$

Step 4: Determine the Absolute Maximum and Minimum

Now we have the function values at the endpoints and the critical point:

• $f(-6) = 16$
• $f(-2) = 0$
• $f(6) = 16$

The absolute maximum value of $f(x)$ on the interval $[-6, 6]$ is $16$, occurring at $x = -6$ and $x = 6$. The absolute minimum value of $f(x)$ on this interval is $0$, occurring at $x = -2$.

Final Answer

The correct answer is:

• The absolute maximum is $16$ at $x = -6$ and $x = 6$.
• The absolute minimum is $0$ at $x = -2$.

Let me know if you want more details or have any questions!

Related Questions:

1. What is the importance of checking the endpoints in determining absolute extrema?
2. How do we interpret fractional exponents like $\frac{4}{3}$ in calculus?
3. What is the difference between absolute and local extrema?
4. Why does setting the derivative to zero help in finding critical points?
5. How do you handle negative bases with fractional exponents?

Tip: Always check both endpoints and critical points to ensure you capture all absolute extrema on a closed interval.