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

1 answer

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 \( x_1 = 1 \):

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

  1. 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.