To determine the absolute extrema of the function $f(x) = x^{1/3} - 5$ on the interval $[-4, 3]$, we need to evaluate $f(x)$ at the critical points within the interval as well as at the endpoints.

Step-by-Step Solution

1. Find the derivative of $f(x)$ to determine critical points: $f(x) = x^{1/3} - 5$ Using the power rule, the derivative $f'(x)$ is: $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$

2. Set the derivative equal to zero to find critical points: $\frac{1}{3x^{2/3}} = 0$ This equation has no solutions since the derivative does not equal zero for any real $x$. However, $f'(x)$ is undefined at $x = 0$, so $x = 0$ is a critical point.

3. Evaluate $f(x)$ at the endpoints and at the critical point $x = 0$:

   • At $x = -4$: $f(-4) = (-4)^{1/3} - 5 = -\sqrt[3]{4} - 5 \approx -6.5874$
   • At $x = 3$: $f(3) = 3^{1/3} - 5 = \sqrt[3]{3} - 5 \approx -3.1347$
   • At $x = 0$: $f(0) = 0^{1/3} - 5 = -5$

4. Determine the absolute minimum and maximum values:

   • The smallest value is $f(-4) \approx -6.5874$, which occurs at $x = -4$.
   • The largest value is $f(3) \approx -3.1347$, which occurs at $x = 3$.

Final Answer

• Absolute minimum: $f(x) = -6.5874$ at $x = -4$
• Absolute maximum: $f(x) = -3.1347$ at $x = 3$

Would you like more detailed steps or explanations for any part?

1. What would the derivative look like if the function had a different exponent?
2. How does the absolute maximum change if the interval changes?
3. What if the interval was open? Would the result be different?
4. How would we solve this if $f(x)$ was a trigonometric function?
5. Can you find the absolute extrema for a polynomial function in a closed interval?

Tip: Always check both endpoints and critical points within a closed interval for absolute extrema.