# [SOLVED] average value

• Jun 11th 2008, 09:57 PM
chukie
[SOLVED] average value
Let A represent the average value fo the function f(x) on the interval [0,6]. Is there a value of c for which the average value of f(x) on the interval [0,c] is greater than A? Why or why not?

http://img179.imageshack.us/img179/2492/graphyu4.th.png

I really hv no idea how to do this problem. Looking at the graph, I thot that there is a value of c because at x=7 the area is greater than x=6? but that's totally just a guess (Worried)
• Jun 11th 2008, 10:04 PM
Mathstud28
Quote:

Originally Posted by chukie
Let A represent the average value fo the function f(x) on the interval [0,6]. Is there a value of c for which the average value of f(x) on the interval [0,c] is greater than A? Why or why not?

http://img179.imageshack.us/img179/2492/graphyu4.th.png

I really hv no idea how to do this problem. Looking at the graph, I thot that there is a value of c because at x=7 the area is greater than x=6? but that's totally just a guess (Worried)

You tell us, average value of a function $f(x)$ on a generalized interval $[a,b]$ is given by

$\frac{1}{b-a}\int_a^{b}f(x)dx$...Now think, graphically will there be a point $\in[a,b]$ such that

$f(c)>\frac{1}{b-a}\int_a^{b}f(x)dx$?
• Jun 11th 2008, 10:39 PM
chukie
Quote:

Originally Posted by Mathstud28
You tell us, average value of a function $f(x)$ on a generalized interval $[a,b]$ is given by

$\frac{1}{b-a}\int_a^{b}f(x)dx$...Now think, graphically will there be a point $\in[a,b]$ such that

$f(c)>\frac{1}{b-a}\int_a^{b}f(x)dx$?

umm no there isnt a value of c? because at x=6 it's already at a maximum? im not sure if im thinking in the right direction
• Jun 12th 2008, 12:24 AM
Reckoner
Quote:

Originally Posted by chukie
umm no there isnt a value of c? because at x=6 it's already at a maximum? im not sure if im thinking in the right direction

Your reasoning is fine. For any $c\neq6$, the average will be smaller since either more points will be below the maximum (for $c > 6$) or the maximum will be lower (for $c < 6$).

You could make this a bit more formal, but I think a simple explanation is probably all that the question is after.