# Math Help - Derivative Inequality

1. ## Derivative Inequality

I'm trying to think this out, but I just can't get it to make any sense. Let f be continuous on [a,b] (a<b) and differentiable on (a,b). If f(a) = 0 and abs(f'(x)) <= c*abs(f(x)) on (a,b), show that abs(f'(x)) <= (c^2)*(x-a)*abs(f(d)), a < d < x. Does anyone know what is needed to be done here? Thanks in advance.

2. I think I can use the Mean Value Theorem in the form f(x) = f'(d)(x-a) where a < d < x. OK, so I get:
abs(f'(x)) <= c*abs(f'(d)(x-a)) = c*(x-a)*abs(f'(d)). What do I do now?

3. Originally Posted by agentZERO
I'm trying to think this out, but I just can't get it to make any sense. Let f be continuous on [a,b] (a<b) and differentiable on (a,b). If f(a) = 0 and abs(f'(x)) <= c*abs(f(x)) on (a,b), show that abs(f'(x)) <= (c^2)*(x-a)*abs(f(d)), a < d < x. Does anyone know what is needed to be done here? Thanks in advance.
I am going to restate a problem a little bit because I usually use the same notation over and over again.

1)Let $f(x)$ be a continous function on $[a,b]$ differenciable on $(a,b)$.
2)Let $f(a)=0$.
3)There exists $M\in \mathbb{R}^+$ so that $|f'(x)|\leq M|f(x)|$ for all $x\in (a,b)$.

To show: For any $x\in (a,b)$ we can choose $c\in (a,x)$ so that $|f'(x)|\leq M^2(x-a)|f(c)|$.

Proof: By #1 use the mean value theorem on $f$ on the interval $[a,x]$ that means $f(x) - f(a) = f'(c)(x-a)$ for some $a. Using #2 we can write $f(x) = f'(c)(x-a)$ thus $|f(x)|\leq |f'(c)|(x-a) \leq M(x-a)$. By #3 we know $|f'(x)|\leq M|f(x)|$ by what we just got it means $|f'(x)|\leq M|f(x)|\leq M|f(c)|(x-a)$. Q.E.D.

4. Thank you very much.

5. I also have to show that
abs(f(x)) <= (c^n)*[(x-a)^n]*M where M = max(abs(f(x))) on [a,b].

I'm going to use my notation:
I use the MVT f(x) = f'(d)(x-a).
Then, abs(f(x)) <= abs(f'(d))*(x-a) <= c*abs(f(d))*(x-a) <= c(x-a). I understand that M will be greater than or equal to all abs(f(x)), but how do I get the c and the (x-a) to the nth power?

6. Alright, I figured the last inequality out. I just need one last thing,
Prove that f = 0 on [a,b] given the conditions at the beginning, namely:
-f continuous on [a,b] (a<b)
-f differentiable on (a,b)
-f(a) = 0
-abs(f'(x)) <= c*abs(f(x)) on (a,b)