How do you find the average rate of change for the function # f (z) = 5 - 8z^2# on the indicated intervals [-8,3]?

Question

Emilia Aguilar · Accepted Answer

40

The #color(blue)"average rate of change"# of f(z)over an interval between 2 points (a ,f(a) and (b ,f(b) is the slope of the #color(blue)"secant line"# connecting the 2 points. to determine the two points' average rate of change. #color(red)(|bar(ul(color(white)(a/a)color(black)((f(b)-f(a))/(b-a))color(white)(a/a)|)))# Here, b = 3 and a = - 8. #rArrf(-8)=5-8(-8)^2=-507# and #f(3)=5-8(3)^2=-67# The mean rate of variation between (3,-67) and (-8,-507) is #(-67-(-507))/(3-(-8))=(440)/(11)=40# Accordingly, the average of all the tangent line slopes to the f(z) graph between (-8,-507) and (3,-67) is 40.

Isabella Ackerman · Answer

undefined To find the average rate of change for the function $ f(z) = 5 - 8z^2 $ on the interval $[-8,3]$, use the formula:

$$ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} $$

where $a$ and $b$ are the endpoints of the interval.

Plug in the values:

$$ f(-8) = 5 - 8(-8)^2 = 5 - 8(64) = 5 - 512 = -507 $$

$$ f(3) = 5 - 8(3)^2 = 5 - 8(9) = 5 - 72 = -67 $$

$$ \text{Average Rate of Change} = \frac{(-67) - (-507)}{3 - (-8)} = \frac{-67 + 507}{3 + 8} = \frac{440}{11} = 40 $$

Therefore, the average rate of change for the function on the interval $[-8,3]$ is $40$.