To find the absolute extrema (maximum and minimum values) of the function $f(x) = (x - 1)^{\frac{4}{3}}$ on the interval $[-7, 7]$, we need to follow these steps:

Steps

1. Find the derivative $f'(x)$ to locate any critical points within the interval.
2. Evaluate $f(x)$ at the critical points as well as the endpoints of the interval, $x = -7$ and $x = 7$.
3. Compare the values to determine the absolute maximum and minimum.

Let's go through each step in detail.

Step 1: Differentiate $f(x)$

The function is $f(x) = (x - 1)^{\frac{4}{3}}$. We apply the power rule:

$f'(x) = \frac{4}{3}(x - 1)^{\frac{4}{3} - 1} \cdot 1 = \frac{4}{3}(x - 1)^{\frac{1}{3}}$

This derivative is undefined when $x = 1$ because $(x - 1)^{\frac{1}{3}}$ has an undefined slope at that point (it's a cusp). This makes $x = 1$ a critical point.

Step 2: Evaluate $f(x)$ at Critical Points and Endpoints

Now, we evaluate $f(x)$ at:

• The endpoints $x = -7$ and $x = 7$.
• The critical point $x = 1$.

1. At $x = -7$: $f(-7) = (-7 - 1)^{\frac{4}{3}} = (-8)^{\frac{4}{3}}$ To evaluate $(-8)^{\frac{4}{3}}$, we can rewrite it as: $(-8)^{\frac{4}{3}} = \left( (-8)^{\frac{1}{3}} \right)^4 = (-2)^4 = 16$ So, $f(-7) = 16$.

2. At $x = 7$: $f(7) = (7 - 1)^{\frac{4}{3}} = (6)^{\frac{4}{3}}$ Calculating $6^{\frac{4}{3}}$: $6^{\frac{4}{3}} = \left( 6^{\frac{1}{3}} \right)^4$ Approximating, $6^{\frac{4}{3}} \approx 15.59$.

3. At $x = 1$: $f(1) = (1 - 1)^{\frac{4}{3}} = 0$

Step 3: Determine the Absolute Extrema

The values of $f(x)$ at these points are:

• $f(-7) = 16$
• $f(7) \approx 15.59$
• $f(1) = 0$

So:

• The absolute maximum of $f(x)$ on $[-7, 7]$ is 16, occurring at $x = -7$.
• The absolute minimum of $f(x)$ on $[-7, 7]$ is 0, occurring at $x = 1$.

Answer

• Absolute Maximum: $16$ at $x = -7$
• Absolute Minimum: $0$ at $x = 1$

Would you like more details on any specific step?

Here are 5 related questions to explore this topic further:

1. What are the general steps to finding absolute extrema on a closed interval?
2. How do we handle derivatives when they involve fractional exponents?
3. What is the significance of critical points that occur due to undefined derivatives?
4. Why does $f(x) = (x - 1)^{\frac{4}{3}}$ have a cusp at $x = 1$?
5. How do we approximate values like $6^{\frac{4}{3}}$ without a calculator?

Tip: For functions involving fractional exponents, be cautious with points where the derivative is undefined, as they often indicate special features like cusps or vertical tangents.