Question

Given f(x) = x2 + 2x + 9, find the average rate of change of f(x) over each of the following pairs of intervals. (a) [1.9, 2] and [1.99, 2] average rate of change over [1.9, 2] 5.9 Correct: Your answer is correct. average rate of change over [1.99, 2] 5.99 Correct: Your answer is correct. (b) [2, 2.1] and [2, 2.01] average rate of change over [2, 2.1] 6.1 Correct: Your answer is correct. average rate of change over [2, 2.01] 6.01 Correct: Your answer is correct. (c) What do the calculations in parts (a) and (b) suggest the instantaneous rate of change of f(x) at x = 2 might be?

Answer 1

Answer:

Required average rate of change over the interval [1.9, 2] is 5.9, [1.99, 2] is 5.99, [2, 2.1] is 6.1, [2, 2.01] is 6.01 and the instantaneous change at x=2 is 6.

Step-by-step explanation:

Given function is,

$f(x)=x^2+2x+9$

To find the avarage rate of change over given intervals. We know from Lagranges Mean value theorem, the average rate of change of a function F(x) over a interval $a\leq x\leq b$ is, $\frac{f(b)-f(a)}{b-a}$.

(a) On the interval,

• [1.9, 2]

$\frac{f(2)-f(1.9)}{2-1.9}= \frac{17-16.41}{0.1}=5.9$

• [1.99, 2]

$\frac{f(2)-f(1.99)}{2-1.99}= \frac{17-16.9401}{0.1}=5.99$

(b) On the interval,

• [2, 2.1]

$\frac{f(2.1)-f(2)}{2.1-2}= \frac{17.61-17}{0.1}=6.1$

• [2, 2.01]

$\frac{f(2.01)-f(2)}{2.01-2}= \frac{17.0601-17}{0.01}=6.01$

(c) Instantaneous rate of change at x=2 is,

$\lim_{x\to 2}\frac{\Delta y}{\Delta x}=\lim_{x\to 2}\frac{f(2)-f(x)}{2-x}$

$=\lim_{x\to 2}\frac{17-x^2-2x-9}{2-x}$

$=\lim_{x\to 2}\frac{-(x+4)(x-2)}{-(x-2)}$

$=\lim_{x\to 2}(x-4)$

$=6$

Hence the results.