To plot five points for the function $f(x) = \sqrt[3]{x} + 6$, we will choose five values of $x$ and compute the corresponding values of $f(x)$.

Let's compute for $x = -8, -1, 0, 1, 8$:

1. When $x = -8$: $f(-8) = \sqrt[3]{-8} + 6 = -2 + 6 = 4$

2. When $x = -1$: $f(-1) = \sqrt[3]{-1} + 6 = -1 + 6 = 5$

3. When $x = 0$: $f(0) = \sqrt[3]{0} + 6 = 0 + 6 = 6$

4. When $x = 1$: $f(1) = \sqrt[3]{1} + 6 = 1 + 6 = 7$

5. When $x = 8$: $f(8) = \sqrt[3]{8} + 6 = 2 + 6 = 8$

Thus, the five points are:
$(-8, 4), (-1, 5), (0, 6), (1, 7), (8, 8)$

I'll now plot these points for you.

matplotlib.pyplot as plt import numpy as np

Values of x and corresponding f(x)

x_values = np.array([-8, -1, 0, 1, 8]) y_values = np.cbrt(x_values) + 6

Plotting the points

plt.figure(figsize=(6,6)) plt.plot(x_values, y_values, 'bo', label="Points") plt.plot(x_values, y_values, 'r-', label="Curve f(x) = ^3√x + 6") plt.title("Plot of f(x) = ^3√x + 6") plt.xlabel("x") plt.ylabel("f(x)") plt.axhline(0, color='black',linewidth=0.5) plt.axvline(0, color='black',linewidth=0.5) plt.grid(True) plt.legend() plt.show()Here is the plot for the function $f(x) = \sqrt[3]{x} + 6$ with the five points:
$(-8, 4), (-1, 5), (0, 6), (1, 7), (8, 8)$

Would you like more details on how the graph behaves, or do you have any further questions?

Related Questions:

1. What happens to $f(x)$ as $x$ approaches very large or very small values?
2. How does the graph of $f(x) = \sqrt[3]{x} + 6$ compare to the graph of $f(x) = \sqrt{x} + 6$?
3. Can you calculate the slope of the tangent line at one of the points (e.g., at $x = 0$)?
4. How would the graph change if we added a constant to $x$ (e.g., $f(x) = \sqrt[3]{x + 2} + 6$)?
5. What is the inverse of the function $f(x) = \sqrt[3]{x} + 6$?

Tip: The cube root function is continuous and smooth across all real numbers, unlike square roots which are restricted to non-negative values of $x$.