Question

A function is given. h(t) = t2 + 3t; t = −1, t = 2

Answer 1

Answer:

Step-by-step explanation:

h(t)=t ^2+3t  

t= -1, t=2

(a) Net Change of the function between given values  

h(t)=t ^2+3t  

h(-1)=( -1)^2+3(-1)

       = 1+(-3)

       =-2

h(2)=( 2)^2+3(2)

       = 4+(6)

       =10

net change =h(2) - h(-1)  

                   =10 -(-2)

                    =12

b, average of change  

net change divided by the change in the x- values  

12/(2-(-1) )

12/(2+1)

12/3= 6  

Average of change =6

Answer 2

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 $f(x)$ over the interval $a\leq x\leq b$ is given by this expression:

average rate of change =  $\frac{net \:change \:in \:y}{change \:in \:x} = \frac{f(b)-f(a)}{b-a}$

It is a measure of how much the function changed per unit, on average, over that interval.

Given:

$h(t)=t^2+3t\\\\t=-1\\t=2$

(a) To find the net change of the function, first we calculate the values of $h(2)$ and $h(-1)$

$h(-1)=(-1)^2+3(-1)=-2\\\\h(2)=(2)^2+3(2)=10$

The net change is simply the difference

$h(2)-h(-1)=10-(-2)=12$

(b) The average rate of change takes the net change and divides it by the change in the $t$ value.

$\frac{12}{2-(-1)} =\frac{12}{3}=4$