Question

Mean value Theorum

Answer 1

So I've got the average of 8/3, but how do I find the point?

Answer 2

Which problem ?

Answer 3

Sorry, no 10

Answer 4

I've worked out a point of (64/9, 8/3) but not sure its correct or not

Answer 5

#10, we have an odd function. So the average value over any symmetric interval (-a, a) must be 0 right ?

Answer 6

I suggest you graph the function and see

Answer 7

\[Average Value = \frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x)\]

Answer 8

Honestly, I could but I wouldnt know what to look for.

Answer 9

Look at the graph :

https://www.desmos.com/calculator/b8dqvxb6uj

Answer 10

Notice that the curve is below x axis in the interval [-8, 0]
and it is above x axis in the interval [0, 8]

Since the area is same on either side, the overall area adds up to 0.

Answer 11

Yea but the bounds are set to [0, 16] here

Answer 12

\[\begin{align}Average Value &= \frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x) \,dx\\~\\

&=\frac{ 1 }{ 8--8 }\int\limits_{-8}^{8}2x^{1/3}\,dx\\~\\

&=\frac{ 1 }{ 8--8 }*0\\~\\
&=0

\end{align}\]

Answer 13

Oh sorry, I was looking at 9th problem.

Answer 14

no worries lol, I wasnt sure if it was you or me

Answer 15

Okay 8/3 is correct for the average value

Lets use MVT to find the point at which this occurs

Answer 16

so set f(x)=8/3 right?

Answer 17

One moment pls

Answer 18

sure

Answer 19

Yeah you're right. Just need to solve
\[f(c) = 8/3\]

Answer 20

\[\sqrt{c} = 8/3\]

Answer 21

great thanks!