Question

Let f be defined as follows.

y = f(x) = x ^ 2 - 7x
(a) Find the average rate of change of y with respect to x in the following intervals.
from x = 6 to x = 7____
from x = 6 tox = 6.5___
from x = 6 to x = 6.1____
(b) Find the (instantaneous) rate of change of y at x = 6.

Answer 1

(a)Average Rate of Change = (\frac{f(6.1) - f(6)}{6.1 - 6} = \frac{(6.1^2 - 7 \cdot 6.1) - (6^2 - 7 \cdot 6)}{0.1} \approx -5.39)

(b) 5

Explanation:

(a) Average rate of change from (x = 6) to (x = 7): Average Rate of Change = (\frac{f(7) - f(6)}{7 - 6} = \frac{(7^2 - 7 \cdot 7) - (6^2 - 7 \cdot 6)}{1} = \frac{7 - 42 - 6 + 42}{1} = -1)Average rate of change from (x = 6) to (x = 6.5): Average Rate of Change = (\frac{f(6.5) - f(6)}{6.5 - 6} = \frac{(6.5^2 - 7 \cdot 6.5) - (6^2 - 7 \cdot 6)}{0.5} \approx -5.75)Average rate of change from (x = 6) to (x = 6.1): Average Rate of Change = (\frac{f(6.1) - f(6)}{6.1 - 6} = \frac{(6.1^2 - 7 \cdot 6.1) - (6^2 - 7 \cdot 6)}{0.1} \approx -5.39)

(b) Instantaneous rate of change at (x = 6): To find the instantaneous rate of change at (x = 6), we can calculate the derivative of (f(x)) and then evaluate it at (x = 6).(f(x) = x^2 - 7x) (f'(x) = 2x - 7)Instantaneous Rate of Change at (x = 6) = (f'(6) = 2 \cdot 6 - 7 = 12 - 7 = 5)So, the answers are: (a)From (x = 6) to (x = 7): -1From (x = 6) to (x = 6.5): approximately -5.75From (x = 6) to (x = 6.1): approximately -5.39(b) The instantaneous rate of change at (x = 6) is 5.