# Find c in (a, b) so tangent passes through 0

• Apr 8th 2010, 06:38 PM
Raied
Find c in (a, b) so tangent passes through 0
Let a; b be real numbers such that 0 < a < b. Let f be a function continuous on [a, b]
and dif erentiable on (a, b). Assume that f satis es the property
f(a) = f(b) = 0 and f'(a) = 0:
Show that there exist c in (a; b) such that the tangent to f at c passes through 0.

Any help? I am totaly lost.
• Apr 8th 2010, 06:39 PM
dwsmith
Sounds like the Mean Value Theorem
• Apr 9th 2010, 05:19 AM
Raied
hmm or could it be Rolles Theorem...
Anyon else?
• Apr 9th 2010, 05:48 AM
dwsmith
When you wrote $\displaystyle f'(a)=0$, was that supposed to be $\displaystyle f'(c)=0$

If you meant $\displaystyle f'(c)=0$, then it is Mean Value.

It is just different to see $\displaystyle f'(a)=0$ since that doesn't come up in either theorem.
• Apr 9th 2010, 06:18 AM
Raied
No it is actually f'(a)...
That is why I am getting confused. I have never seen that before
• Apr 9th 2010, 06:38 AM
Plato
The tangent line is $\displaystyle y=f'(c)x-cf'(c)+f(c)$.
To pass through (0,0) we must have $\displaystyle f(c)-cf'(c)=0$.
Consider the function $\displaystyle \frac{f(x)}{x}$ apply Rolle's Theorem.
• Apr 9th 2010, 07:13 AM
Raied
is x = a?
• Apr 9th 2010, 07:52 AM
Plato
Suppose that $\displaystyle g(x)=\frac{f(x)}{x}$ then $\displaystyle g(a)=g(b)=0$.
Moreover, on $\displaystyle (a,b)$ $\displaystyle g'(x) = \frac{{xf'(x) - f(x)}}{{x^2 }}$ so if $\displaystyle g'(c)=0$ then $\displaystyle cf'(c) - f(c)=0$.