# How do you find the average and instantaneous rate of change of y=2x^2-2 over the interval [1,3/2]?

Nov 7, 2017

Use power rule to find derivative, and average max and minimum vals for derivative to find avg. $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 x , \overline{\frac{\mathrm{dy}}{\mathrm{dx}}} = \overline{y} ' = 5$

#### Explanation:

The instantaneous rate of change will be calculated at any point on the interval, by inserting the relevant x value into the derivative. So, the instantaneous rate of change at 1 is the derivative at x=1.

The average rate of change will be the sum of the maximum and minimum values the derivative takes on the interval, divided by 2.

Our derivative will be found via the power rule which states thatnfor $f \left(x\right) = a {x}^{n} + b {x}^{n - 1} + \ldots + m x + k , \frac{\mathrm{df}}{\mathrm{dx}} = n a {x}^{n - 1} + \left(n - 1\right) {x}^{n - 2} + \ldots + m$

Thus for $y = 2 {x}^{2} - 2 , \frac{\mathrm{dy}}{\mathrm{dx}} = 4 x$.

The instantaneous rate of change at any rate along the interval then, is calculated by $y ' = 4 x$. Since this is linear rather than quadratic, and because it increases with x, we can find our average rate of change easily:

$\overline{y} ' = \frac{y ' \left(\frac{3}{2}\right) + y ' \left(1\right)}{2} = \frac{6 + 4}{2} = 5$