Question

What’s the difference between the rates of change in the cube root y = cuberoot x + 2? Round to nearest tenth

Answer 1

To find the difference between the rates of change in the function \( y = \sqrt[3]{x} + 2 \), we need to first determine the derivative of the function, which will give us the rate of change at any point \( x \).

The derivative of \( y = \sqrt[3]{x} + 2 \) can be computed as follows:

1. The derivative of \( \sqrt[3]{x} \) is \( \frac{1}{3}x^{-\frac{2}{3}} \).
2. The derivative of the constant 2 is 0.

Thus, the derivative \( y' \) is:

\[
y' = \frac{d}{dx}(\sqrt[3]{x} + 2) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3\sqrt[3]{x^2}}
\]

This derivative gives us the rate of change of \( y \) with respect to \( x \).

Next, we'll compute the rate of change at two different points. Let's choose \( x_1 = 1 \) and \( x_2 = 8 \).

1. For \( x_1 = 1 \):

\[
y'(1) = \frac{1}{3\sqrt[3]{1^2}} = \frac{1}{3 \cdot 1} = \frac{1}{3} \approx 0.33
\]

2. For \( x_2 = 8 \):

\[
y'(8) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3\sqrt[3]{64}} = \frac{1}{3 \cdot 4} = \frac{1}{12} \approx 0.0833
\]

Now we calculate the difference between these rates of change:

\[
\text{Difference} = y'(1) - y'(8) = \frac{1}{3} - \frac{1}{12}
\]

To perform this subtraction, we need a common denominator. The common denominator of 3 and 12 is 12.

\[
\frac{1}{3} = \frac{4}{12}
\]

So,

\[
\text{Difference} = \frac{4}{12} - \frac{1}{12} = \frac{3}{12} = \frac{1}{4} = 0.25
\]

Finally, rounding to the nearest tenth:

\[
\text{Final Answer} = 0.3
\]

Thus, the difference between the rates of change at \( x = 1 \) and \( x = 8 \) rounded to the nearest tenth is **0.3**.