Use the Mean Value Theorem to show that:

a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

b) Assume that |f ' (x)| < or equal to C < 1 for all x. Show that f (x) = x has at most one solution.

Please someone help me with this Mean Value Theorem Proof. Thanks

2. Use the Mean Value Theorem to show that:

a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).
Suppose f is not one-one on the interval then there exists u, v in (a,b) such that f(u)=f(v).
Then by the mean value theorem there exists a point w in (u,v) such that f'(w)=0, a contradiction.

RonL

3. I have a test in a couple of hours. I can't state part a with a contradiction and I need a full blown proof if someone can help me with that.

Also for part b, the assumption that f is differentiable on
(a,b) should have been explicitly stated. If someone can help me finish these 2 parts off into one giant proof , it would be great appreciated. Thank you

4. If anyone can still help me with this, it would be greatly appreciated. Thanks

5. Surely you have been helped with this.
The above posting is definitive in this question.
What more do you want?

6. Originally Posted by Swamifez
Use the Mean Value Theorem to show that:

a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

b) Assume that |f ' (x)| < or equal to C < 1 for all x. Show that f (x) = x has at most one solution.

b) consider g(x)=f(x)-x, then g'(x) != 0 on any interval [-u,u], since:

|g'(x)|=|f(x)-1|>=||f(x)|-1| != 0.

Hence by a) g(x) is one-one on [-u,u], so g(x)=0 has at most one root on
such an interval, and as u is arbitary at most one root in R.

RonL

7. First, thank you for explaining what you are doing with this sequence of questions. It would have been helpful to know that up front.

Now for another way to do part (b) which can be done as quite separate from part (a). Assume that there are points c & d in [a,b] such that f(c)=c & f(d)=d and c is not d. Using the mean value theorem $\exists e \in [a,b]\left[ {f'(e) = \frac{{f(d) - f(c)}}{{d - c}} = 1} \right].$
But $\left| {f'(e)} \right| \le C < 1$ which is a contradiction.
Thus there is a most one x in [a,b] such that f(x)=x.