Over which of the following intervals does f(x)= −1/x have the greatest average rate of change?
a) [2, -1/4]
b) [2, 5]
c) [1, 5/2]
d) [-4, 2]
find f(x) for the values at the ends of the interval
divide the difference in the f(x) values by the difference in the x values
You know that the graph has an asypmote at x=0, so intervals closer to the asymptote have the greatest slope. Since all of these intervals are relatively narrow, I pick C even without doing the calculations.
A and D automatically are out, since they include the asymptote, so the average rate of change is undefined.
To find the interval over which the function f(x) = -1/x has the greatest average rate of change, we need to calculate the average rate of change for each interval and compare them.
The average rate of change of a function over a given interval is given by the formula:
Average rate of change = (f(b) - f(a))/(b - a)
Let's calculate the average rate of change for each interval:
a) [2, -1/4]
Average rate of change = (f(-1/4) - f(2))/(-1/4 - 2)
= (-4 - (-1/2))/(-1/4 - 2)
= (-4 + 1/2)/(-1/4 - 2)
= (-7/2)/(1/4 - 8/4)
= (-7/2)/(-7/4)
= (-7/2)*(4/7)
= -14/7
= -2
b) [2, 5]
Average rate of change = (f(5) - f(2))/(5 - 2)
= (-1/5 - (-1/2))/(5 - 2)
= (-2/10 + 5/10)/(3)
= (3/10)/(3)
= 3/10 * 1/3
= 3/30
= 1/10
c) [1, 5/2]
Average rate of change = (f(5/2) - f(1))/(5/2 - 1)
= (-2/5 - (-1))/(5/2 - 1)
= (-2/5 + 1)/(5/2 - 1)
= (-2/5 + 1)/(5/2 - 2/2)
= (-2/5 + 1)/(5/2 - 2/2)
= (-2/5 + 1)/(5/2 - 2/2)
= (-2/5 + 1)/(5/2 - 1/2)
= (-2/5 + 1)/(5/2 - 1/2)
= (-2/5 + 1)/(5/2 - 1/2)
= (-2/5 + 1)/(5/2 - 1/2)
= (-7/10)/(4/2)
= (-7/10)/(2)
= -7/20
d) [-4, 2]
Average rate of change = (f(2) - f(-4))/(2 - (-4))
= (-1/2 - (-1/(-4)))/(2 - (-4))
= (-1/2 - (+1/4))/(2 + 4)
= (-2/4 - (+1/4))/(6)
= (-3/4)/(6)
= -3/24
Comparing the average rates of change for each interval, we can see that the interval with the greatest average rate of change is:
a) [2, -1/4], with an average rate of change of -2.
Therefore, the correct answer is option a) [2, -1/4].