Posted by Someone on .
if f(x)= (x^29)(x^2+1) how many numbers in the interval [1,1] satisfy the mean value theorem?
None
1
2
3
Will someone please explain in detail?

Calculus 
Steve,
absolute value problems can get tricky, because the functions are not differentiable at any cusps.
However, since (x^29)(x^2+1) is negative on [1,1],
f(x) = (x^29)(x^2+1)
This is a nice smooth polynomial, so it satisfies the requirements of the MVT on [1,1].
Now take a look at the graph:
http://www.wolframalpha.com/input/?i=%7C(x%5E29)(x%5E2%2B1)%7C
Since f(1) = f(1) we see that there is a single point where f'(c) = 0 on [1,1].
To show that algebraically, we just go through the steps:
(f(1)f(1))/2 = 0
f'(x) = 4x(4x^2)
f'(0) = 0 
Calculus 
Ke$ha,
so the answer is none?

Calculus 
Steve,
Excuse me?
Did you look at the graph?
Did you actually read what I wrote?
I don't mind helping you arrive at a solution, but I do expect you to actually read it, rather than just hold out your hand and say "gimme the answer." In fact, I did give you the answer. 
Calculus 
Ke$ha,
Yes I read it I still don't get it and 0 isn't an answer choice.

Calculus 
Steve,
That is because x=0 is the single point where f'(x) = 0
the number zero is a single choice, not an absence of choices!
The answer is 1
There is 1 (one) point on the interval [1,1] where
f(1)f(1)
 = 0
1  (1)
That point is at x=0.
Just take a deep breath and read what is written. 
Calculus 
Ke$ha,
wait its 3 right?

Calculus 
Ke$ha,
oh ok thank you!!!! sorry