be f(x) Continuous function in [0,2] , and Differentiable in (0,2).
0<f(1)<f(0)<f(2).
Prove that the Differentiable become zero in any point in the open Interval (0,2).
thanks
Is it included in the hypothesis $\displaystyle f'$ continuous in $\displaystyle (0,2)$ ? . In such a case, you only need to apply Lagrange's theorem to the intervals $\displaystyle [0,1]$ and $\displaystyle [1,2]$ .
Regards.
Fernando Revilla
I think what you want is a proof that the derivative is 0 at some point in (0, 2). "Any point" would mean every point and that is not true.
Use the "mean value theorem" on the intervals (0, 1) and (1, 2). There exist a number, a, between 0 and 1, such that (f(1)- f(0))/(1- 0)= f(1)- f(0)= f'(a). Since f(0)> f(1), f'(a) is negative. There exist a number, b, between 1 and 2, such that (f(2)- f(1))/(2- 1)= f(2)- f(1)= f'(b). Since f(1)< f(2), f'(b) is positive.
Now, do you have the theorem that says that, though the derivative of a given function is not necessarily continuous, it still must satisfy the "intermediate value property"? If so that immediately tells you that there is some point, c, between a and b and so between 0 and 2, such that f'(c)= 0.
If you do not have that theorem, you should prove it! Define g(x) on [0, 2] by g(x)= (f(x)- f(0))/x as long as x is not 0 and g(0)= f'(0). You can show that g is continous on [0, 2] and so the intermediate value property applies to g: g(x) takes on all values between g(0)= f'(0) and g(2)= (f(2)- f(0))/2. Define h(x) on [0, 2] by h(x)= (f(2)- f(x))/(2- x) as long as x is not 2 and h(2)= f'(2). Then h(x) takes on all values between h(0)= (f(2)- f(0))/2 and h(2)= f'(2). But by the mean value theorem, f'(x) takes on all of those values itself.
This does NOT belong in "pre-calculus"!
Excellent comment HallsofIvy . For that reason I asked about continuity of $\displaystyle f'$ .
Regards.
Fernando Revilla