Question

calc help!

Answer 1

that's not very descriptive :/

Answer 2

it helps if one knows what may be
the issue is, some folks likely do know
but an ambiguous posting message, doesn't indicate them so

Answer 3

i got Yes, x=1 @DanJS @Loser66

Answer 4

Why is it x = 1? You need to show work, as I've told you many times before.

Answer 5

What did you do to get that value?

Answer 6

What is the Mean Value Theorem? I mean what it say?

Answer 7

Look at the bottom of that calc sheet page 5 i linked to you.. it shows the definition there

Answer 8

The point x=# , will be the average value of the function on the given interval [0,root3]

Answer 9

what point?

Answer 10

The mean value theorem says that if you have a continuous function over an interval [a,b], that there will be a value x=c, in the interval, where f(c) will be the average value of the function f(x) over the interval

Answer 11

oh yeah i have the definition

Answer 12

remember a couple probs back, we calculated the average value by evaluating an integral?

Answer 13

yeah the like first question

Answer 14

at a value x=a, the function value f(a) will be the average value in that shaded region.

Answer 15

integrate the function first... what is that?
\[\int\limits [3-x^2]dx = \]

Answer 16

3x-x^3/3

Answer 17

right, now evaluate the definite integral over the integral [0 root3]

Answer 18

The average value f(c) will be
\[f(c) = \frac{ 1 }{ \sqrt{3}-0 }\int\limits_{0}^{\sqrt{3}}[3-x^2]dx\]

Answer 19

and do i solve that?

Answer 20

basically what this means is that there is a value c, in between those two end points [a,b], where the tangent line will have the same slope as the secant line connecting the points a and b...