Respuesta :
Answer:
(a) The net change of the function is 12.
(b) The average rate of change of the function 4.
Step-by-step explanation:
The average rate of change of function [tex]f(x)[/tex] over the interval [tex]a\leq x\leq b[/tex] is given by this expression:
average rate of change = [tex]\frac{net \:change \:in \:y}{change \:in \:x} = \frac{f(b)-f(a)}{b-a}[/tex]
It is a measure of how much the function changed per unit, on average, over that interval.
Given:
[tex]h(t)=t^2+3t\\\\t=-1\\t=2[/tex]
(a) To find the net change of the function, first we calculate the values of [tex]h(2)[/tex] and [tex]h(-1)[/tex]
[tex]h(-1)=(-1)^2+3(-1)=-2\\\\h(2)=(2)^2+3(2)=10[/tex]
The net change is simply the difference
[tex]h(2)-h(-1)=10-(-2)=12[/tex]
(b) The average rate of change takes the net change and divides it by the change in the [tex]t[/tex] value.
[tex]\frac{12}{2-(-1)} =\frac{12}{3}=4[/tex]
Functions can be represented using equations.
- The net difference between both function values is 12
- The average rate of change is 12
The function is given as:
[tex]\mathbf{h(t) = t^2 + 3t}[/tex]
(a) Net change at t = -1 and t = 2
First, we calculate h(-1) and h(2)
[tex]\mathbf{h(-1) = (-1)^2 + 3(-1)}[/tex]
[tex]\mathbf{h(-1) = 1 - 3}[/tex]
[tex]\mathbf{h(-1) = - 2}[/tex]
[tex]\mathbf{h(2) = (2)^2 + 3(2)}[/tex]
[tex]\mathbf{h(2) = 4 + 6}[/tex]
[tex]\mathbf{h(2) = 10}[/tex]
So, the net difference between both is:
[tex]\mathbf{Net = |h(2) - h(-1)|}[/tex]
[tex]\mathbf{Net = |10--2|}[/tex]
[tex]\mathbf{Net = |10+2|}[/tex]
[tex]\mathbf{Net = |12|}[/tex]
[tex]\mathbf{Net = 12}[/tex]
Hence, the net difference between both function values is 12
(b) The average rate of change (m)
This is calculated as:
[tex]\mathbf{m = \frac{h(2) - h(-1)}{2--1}}[/tex]
[tex]\mathbf{m = \frac{10--2}{2--1}}[/tex]
[tex]\mathbf{m = \frac{12}{1}}[/tex]
[tex]\mathbf{m =12}[/tex]
Hence, the average rate of change is 12
Read more about average rates of change at:
https://brainly.com/question/23715190
