Question

Consider x^2 on an interval [0,1/2]. The Mean Value theorem suggest that there is a number c in (0.1/1/2) such that f'(c) is equal to a particular d.What is d?

Answer 1

find the slope of the secant line that passes through (.1, .01) and (.5, .25)

Answer 2

sorry, my browser crashed

Answer 3

is the interval [0, 1/2] or [0.1, 1/2]?

Answer 4

well, I'll assume [0, 1/2] then. Let's say f(x) = x^2

Answer 5

Then the slope of the line between (0, f(0) and (.5, f(.5)) is
(.25-0)/(.5-0) = 0.5

Answer 6

next find where the derivative equals 0.5. This is the point where the tangent line is parallel to the secant line between (0, f(0) and (.5, f(.5))

Answer 7

sorry I crashed too haha

Answer 8

so f'(x) = 2x = 0.5
x = 0.25

Answer 9

so is d 0.25?

Answer 10

yes

Answer 11

well, no actually

Answer 12

d is 0.5

Answer 13

oh k, that makes more sense

Answer 14

because f'(0.25) = 0.5
so c = 0.25
d = 0.5

Answer 15

thanks a lot

Answer 16

no problem