# How do you find the average rate of change of f(x)=3x-2x^2 over interval [-3,4]?

Mar 20, 2016

So mean rate of change is $\frac{15 - 13}{2} = + 1$

#### Explanation:

As you use the term $f \left(x\right)$ I am assuming you are using early stage Calculus.

Set $y = 3 x - 2 {x}^{2}$ ........................(1)

Increment $x$ by the minute amount of $\delta x$
The this will cause a minute change in y of $\delta y$

So

$y + \delta y = 3 \left(x + \delta x\right) - 2 {\left(x + \delta x\right)}^{2}$

$\implies y + \delta y = 3 x + 3 \delta x - 2 \left({x}^{2} + 2 x \delta x + {\left(\delta x\right)}^{2}\right)$

=> y+deltay=3x+3deltax-2x^2-4xdeltax-2(deltax)^2)

$\implies y + \delta y = 3 x + 3 \delta x - 2 {x}^{2} - 4 x \delta x - 2 {\left(\delta x\right)}^{2}$....(2)

Subtract equation (1) from equation (2)

$\text{ } y + \delta y = 3 x + 3 \delta x - 2 {x}^{2} - 4 x \delta x - 2 {\left(\delta x\right)}^{2}$....(2)
" "underline( y" "= 3x" " -2x^2color(white)(........................) -) ..(1)
$\text{ " deltay=0" } + 3 \delta x + 0 - 4 x \delta x - 2 {\left(\delta x\right)}^{2}$

Divide throughout by $\delta x$

$\frac{\delta y}{\delta x} = 3 - 4 x - 2 \delta x$

${\lim}_{\delta a \to 0} \frac{\delta y}{\delta x} = 3 - 4 x - {\lim}_{\delta x \to 0} \left(2 \delta x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 - 4 x$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
At $x = - 3 \text{ "->" } \frac{\mathrm{dy}}{\mathrm{dx}} = 3 - 4 \left(- 3\right) = + 15$

At $x = + 4 \text{ "->" } \frac{\mathrm{dy}}{\mathrm{dx}} = 3 - 4 \left(+ 4\right) = - 13$

So mean rate of change is $\frac{15 - 13}{2} = + 1$