Question

find the average rate of change of ( g(x) = 6x^2 + \frac{4}{x^3} ) on the interval -2, 4. question help: video

Explanation:

Answer:

To find the average rate of change of \( g(x) = 6x^2 + \frac{4}{x^3} \) on the interval \([-2, 4]\), we use the formula for the average rate of change:

\[
\text{Average rate of change} = \frac{g(b) - g(a)}{b - a}
\]

where \( a = -2 \) and \( b = 4 \).

First, calculate \( g(-2) \):

\[
g(-2) = 6(-2)^2 + \frac{4}{(-2)^3} = 6(4) + \frac{4}{-8} = 24 - \frac{1}{2} = \frac{48}{2} - \frac{1}{2} = \frac{47}{2}
\]

Next, calculate \( g(4) \):

\[
g(4) = 6(4)^2 + \frac{4}{(4)^3} = 6(16) + \frac{4}{64} = 96 + \frac{1}{16} = \frac{1536}{16} + \frac{1}{16} = \frac{1537}{16}
\]

Now, find the difference \( g(4) - g(-2) \):

\[
g(4) - g(-2) = \frac{1537}{16} - \frac{47}{2} = \frac{1537}{16} - \frac{376}{16} = \frac{1537 - 376}{16} = \frac{1161}{16}
\]

Then, find the difference \( b - a \):

\[
4 - (-2) = 6
\]

Finally, the average rate of change is:

\[
\frac{\frac{1161}{16}}{6} = \frac{1161}{16 \times 6} = \frac{1161}{96} = \frac{387}{32} = 12.09375
\]

So the average rate of change is \(\frac{387}{32}\) (or \(12.09375\)).