Question

suppose that $f(t) = t^2 + 2t - 3$.
(a) what is the average rate of change of $f(t)$ over the interval 1 to 2?
(b) what is the (instantaneous) rate of change of $f(t)$ when $t = 1$?

the average rate of change of $f(t)$ over the interval 1 to 2 is $\boxed{}$.
(simplify your answer.)

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \( f(t) \) over the interval \([a, b]\) is given by \(\frac{f(b)-f(a)}{b - a}\). Here, \( a = 1 \) and \( b = 2 \), and \( f(t)=t^{2}+2t - 3 \).

Step2: Calculate \( f(2) \)

Substitute \( t = 2 \) into \( f(t) \):
\( f(2)=(2)^{2}+2\times(2)-3=4 + 4-3=5 \)

Step3: Calculate \( f(1) \)

Substitute \( t = 1 \) into \( f(t) \):
\( f(1)=(1)^{2}+2\times(1)-3=1 + 2-3=0 \)

Step4: Calculate the average rate of change

Using the formula \(\frac{f(2)-f(1)}{2 - 1}\), substitute \( f(2) = 5 \) and \( f(1)=0 \):
\(\frac{5 - 0}{2 - 1}=\frac{5}{1}=5\)

Answer:

5