Question

please help

Answer 1

@3mar

Answer 2

i believe it to be A...

Answer 3

http://www.sosmath.com/calculus/diff/der11/der11.html

Answer 4

@sunnnystrong
Please If you could help here, I would be grateful!

Answer 5

Many thanks for your respond the call @sunnnystrong

Answer 6

NP Lets see here... haha

Answer 7

Mean value theorem states that: see file

Looking at :
\[f(x)=\left| (x^2-10)(x^2+1) \right|\]
between [0,2.5]
\[f'(x)=\frac{ 2x(x^2-10)(2x^2-9) }{ \left| x^2-10\right| }\]

Answer 8

its 3 right????

Answer 9

---> Basically you want to look at the definition:

\[f'(c)=\frac{ f(b)-f(a) }{ b-a }\]
[0,2.25]
aka [a,b]

Find point see such that :
\[f'(c)=\frac{ f(2.25)-f(0) }{ 2.25-0 }\]
\[f'(c)=\frac{ 18.9 }{ 2.25}\]
\[f'(c)=8.4\]

Answer 10

... more work lol

Answer 11

and what after that @sunnnystrong ?

Answer 12

\[\frac{ 2x(x^2-10)(2x^2-9 )}{ \left| x^2-10 \right| }=8.4\]
...

Answer 13

^^^ Basically we are looking for what values of x in the derivative make that statement true

I graphed this on wolfram -->

Answer 14

so two :D

Answer 15

@iwanttogotostanford

Answer 16

@sunnnystrong

is there something wrong here?

Answer 17

@3mar
Yeah i used wolfram for that derivative sooo 0_0

Answer 18

That's why hardly any work* XD

Answer 19

Anyway....after examination and analysis...

I have found that the derivative of that equation - to satisfies the mean value theorem in the interval [0,2.5] - must intersect the a-axis in three points, just two of them in the interval [0,2.5]....

These two points are \((0.39571,0)\) and \((1.8956,0)\)

So as a conclusion....there are two points in the interval [0,2.5] that make the given function satisfies the mean value theorem....

Answer 20

and special thanks for @sunnnystrong who has initiated us to go through that puzzle!!!!