# Math Help - [SOLVED] average value

1. ## [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?

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

2. 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?

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
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

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

3. 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

4. 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.