Let f be defined as follows.
y=f(x)=x^2-12x
(a) Find the average rate of change of y with respect to x in the following intervals.
from x = 5 to x = 6 ??
from x = 5 to x = 5.5 ??
from x = 5 to x = 5.1 ??
(b) Find the (instantaneous) rate of change of y at x = 5.
??
y(6) = 36 - 72 = -36
y(5) = 25 - 60 = -35
dy/dx = [(-36) - (-35)]/(6-5)
= -1
same way for next 2
eg for x = 5 to 5.1
y(5.1) = -35.19
y(5) = -35
dy/dx = [-35.19 -(-35)]/(5.1-5)
= -1.9
(b) y' = 2x - 12
at x = 5
y' = 10-12 = -2
To find the average rate of change of y with respect to x in a given interval, we need to calculate the difference in y-values over the difference in x-values.
(a) Average rate of change from x = 5 to x = 6:
To find the average rate of change in this interval, we need to calculate f(6) - f(5), divided by 6 - 5.
First, let's evaluate f(6):
f(6) = (6)^2 - 12(6) = 36 - 72 = -36
Next, let's evaluate f(5):
f(5) = (5)^2 - 12(5) = 25 - 60 = -35
Now, let's calculate the average rate of change:
Average rate of change = (f(6) - f(5)) / (6 - 5)
= (-36 - (-35)) / (6 - 5)
= (-36 + 35) / (1)
= -1
Therefore, the average rate of change of y with respect to x in the interval from x = 5 to x = 6 is -1.
(b) Average rate of change from x = 5 to x = 5.5:
Following the same steps as above, let's evaluate f(5.5) and f(5) to find the average rate of change.
f(5.5) = (5.5)^2 - 12(5.5) = 30.25 - 66 = -35.75
Average rate of change = (f(5.5) - f(5)) / (5.5 - 5)
= (-35.75 - (-35)) / (5.5 - 5)
= (-35.75 + 35) / (0.5)
= -0.75 / 0.5
= -1.5
Therefore, the average rate of change of y with respect to x in the interval from x = 5 to x = 5.5 is -1.5.
(c) Average rate of change from x = 5 to x = 5.1:
Following the same steps as above, let's evaluate f(5.1) and f(5) to find the average rate of change.
f(5.1) = (5.1)^2 - 12(5.1) = 26.01 - 61.2 = -35.19
Average rate of change = (f(5.1) - f(5)) / (5.1 - 5)
= (-35.19 - (-35)) / (5.1 - 5)
= (-35.19 + 35) / (0.1)
= -0.19 / 0.1
= -1.9
Therefore, the average rate of change of y with respect to x in the interval from x = 5 to x = 5.1 is -1.9.
(d) Instantaneous rate of change at x = 5:
To find the instantaneous rate of change (also known as the derivative) at x = 5, we need to take the derivative of the function f(x) with respect to x, and then evaluate it at x = 5.
First, let's find the derivative of f(x):
f'(x) = 2x - 12
Now, let's evaluate f'(x) at x = 5:
f'(5) = 2(5) - 12
= 10 - 12
= -2
Therefore, the instantaneous rate of change of y at x = 5 is -2.