Compare the average rates of change for a square root function and a cube root func...

Calculate the average rate of change for $p(x) = \sqrt{x}+4$ and $f(x) = \sqrt[3]{x}+4$ over $[-4, -3]$. For $p(x)$, the average rate of change is $\frac{p(-3)-p(-4)}{-3-(-4)} = \frac{\sqrt{-3}+4 - (\sqrt{-4}+4)}{1}$, which is undefined. However, the graph shows $p(x)$ for $x \ge -4$. Assuming the graph is representative and the interval is meant to be within the domain, let's re-evaluate the problem. The graph shows points at $(-4, 0)$, $(-3, 1)$, $(0, 2)$, $(5, 3)$, $(12, 4)$. The average rate of change for $p(x)$ over $[-4, -3]$ is $\frac{p(-3)-p(-4)}{-3-(-4)} = \frac{1-0}{1} = 1$. For $f(x) = \sqrt[3]{x}+4$, the average rate of change over $[-4, -3]$ is $\frac{f(-3)-f(-4)}{-3-(-4)} = \frac{\sqrt[3]{-3}+4 - (\sqrt[3]{-4}+4)}{1} = \sqrt[3]{-3} - \sqrt[3]{-4} \approx -1.44 - (-1.59) = 0.15$. Thus, the square root function has a greater average rate of change.