Question

Mean Value Theorem question! Will be posted below.

Answer 1

Let
\[f(x)=2-|2x-1|\]

Show that there is no value c such that
\[f(3)-f(0)=f'(c)(3-0)\]

Why does this not contradict the Mean Value Theorem?

Answer 2

so you want to see if f(x) is continuous on [0,3] and differentiable on (0,3)

Answer 3

this is the conditions that must be met before you can apply mean value theorem

Answer 4

or in general
the conditions to be met is:
continuous on [a,b]
differentiable on (a,b)
conclusion if the conditions are met:
there exist c in (a,b) such that f'(c)=(f(b)-f(a))/(b-a)

Answer 5

so anyways
Do you know if f(x)=2-|2x-1| is continuous for all x in [0,3]?

Answer 6

Oh, there's a cusp. So it doesn't satisfy?

Answer 7

yep at 2x-1=0 there is is a sharp turn

Answer 8

you can solve that for x
and see that that value is actually in (0,3)
so the conditions aren't met

Answer 9

Why doesn't it contradict the Mean Value Theorem? <-- I'm not exactly sure what they're asking for here, though.

Answer 10

well did you try to solve the above equation yet?

Answer 11

Is it because there's a value for c when f'(c)=0?

Answer 12

Oh wait no. Hahaha, excuse me, never mind that.

Answer 13

here is a hint:
\[g(x)=|h(x)| \\ g^2(x)=h^2(x) \\ 2 g(x) g'(x)=2 h(x) h'(x) \\ g'(x)=\frac{2 h(x) h'(x)}{2 g(x)} \\ g'(x)=\frac{ h(x) h'(x)}{|h(x)|} \text{ since } g(x)=|h(x)|\]

Answer 14

so the derivative of |2x-1| is?

Answer 15

you can use that formula above

Answer 16

Thank you, I understand now!

Answer 17

\[(|h(x)|)'=\frac{h(x) h'(x)}{|h(x)|}\]


...
well just in case...
I think they want you to solve that one equation first for c and not even think about the mean value theorem
but then they want you to think about how this does or does not go against the mean value theorem

Answer 18

like after the fact

Answer 19

That's what I assumed. It's pretty clear now, thank you! @myininaya

Answer 20

np

Answer 21

you should get x=1/2 as a extraneous solution
and I say extraneous because f'(1/2) doesn't exist as we said before :)

Answer 22

anyways I will stop
you have fun

Answer 23

@myininaya Thank you again! This is really great help.