Question

suppose that ( f(x) = 3x^2 ).
(a) what is the average rate of change of ( f(x) ) over each of the following intervals: 3 to 4, 3 to 3.5, 3 and to 3.1?
(b) what is the (instantaneous) rate of change of ( f(x) ) when ( x = 3 )?

(a) the average rate of change of ( f(x) ) over the interval 3 to 4 is 21.
(simplify your answer.)

(a) the average rate of change of ( f(x) ) over the interval 3 to 3.5 is ( square ).
(simplify your answer.)

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \( f(x) \) over the interval \([a, b]\) is given by \(\frac{f(b) - f(a)}{b - a}\). Here, \( a = 3 \) and \( b = 3.5 \), and \( f(x)=3x^{2} \).

Step2: Calculate \( f(3) \) and \( f(3.5) \)

First, calculate \( f(3) \):
\( f(3)=3\times(3)^{2}=3\times9 = 27 \)
Then, calculate \( f(3.5) \):
\( f(3.5)=3\times(3.5)^{2}=3\times12.25 = 36.75 \)

Step3: Apply the average rate of change formula

Now, use the formula \(\frac{f(3.5)-f(3)}{3.5 - 3}\). Substitute the values of \( f(3.5) \) and \( f(3) \):
\(\frac{36.75 - 27}{0.5}=\frac{9.75}{0.5}=19.5\)

Answer:

\( 19.5 \)