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

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.

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,
`(f(b)-f(a))/(b-a)`.

(a) On the interval,

• [1.9, 2]

`(f(2)-f(1.9))/(2-1.9)= (17-16.41)/(0.1)=5.9`

• [1.99, 2]

`(f(2)-f(1.99))/(2-1.99)= (17-16.9401)/(0.1)=5.99`

(b) On the interval,

• [2, 2.1]

`(f(2.1)-f(2))/(2.1-2)= (17.61-17)/(0.1)=6.1`

• [2, 2.01]

`(f(2.01)-f(2))/(2.01-2)= (17.0601-17)/(0.01)=6.01`

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

`\lim_(x\to 2)(\Delta y)/(\Delta x)=\lim_(x\to 2)(f(2)-f(x))/(2-x)`

`=\lim_(x\to 2)(17-x^2-2x-9)/(2-x)`

`=\lim_(x\to 2)(-(x+4)(x-2))/(-(x-2))`

`=\lim_(x\to 2)(x-4)`

`=6`

Hence the results.