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:
- The derivative of \( \sqrt[3]{x} \) is \( \frac{1}{3}x^{-\frac{2}{3}} \).
- 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 \( x_2 = 8 \).
- For \( x_1 = 1 \):
\[ y'(1) = \frac{1}{3\sqrt[3]{1^2}} = \frac{1}{3 \cdot 1} = \frac{1}{3} \approx 0.33 \]
- 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.
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