differentiability and continuity problem

**maoro** · December 13th 2011, 08:35 AM

#1 Suppose that the differentiable function $f(x)$ satisfies $f(1)=2$ and $f'(x) \leq 1$ . How large can $f(3)$ possibly be?(Done)
If $f(3)=4$ , then show that $f(x)=x+1$ for $1 \leq x \leq 3$ .

How to approach this?

#2 Assume $f(x)$ is continuous on $[a,b]$ , and for any $x\in[a,b]$ , there exists y\in[a,b] such that $|f(y)| \leq \frac{1}{2}|f(x)|$ . Prove that there exists a point $c \in [a,b]$ such that $f(c)=0.$

**FernandoRevilla** · December 13th 2011, 08:51 AM

Originally Posted by maoro

#1 Suppose that the differentiable function $f(x)$ satisfies $f(1)=2$ and $f'(x) \leq 1$ . How large can $f(3)$ possibly be?(Done)
If $f(3)=4$ , then show that $f(x)=x+1$ for $1 \leq x \leq 3$ .

Using the Lagrange Theorem for $f$ on $[1,3]$ we have $f(3)=2f'(\xi)+f(1)$ for some $\xi\in(1,3)$ . If $f(1)=2$ , as $f'(\xi)\leq 1$ we conclude $f(3)\leq 4$ . If $f(3)=4$ then $f'(\xi)=1$ so ... (conclude)

**maoro** · December 13th 2011, 08:58 AM

I have just learned about Mean Value Theorem, is it similar to Lagrange Theorem?
btw, if $f(3)=4$ , we can just show that $f'(\xi)=1$ , but not $f(x)=x+1$

**FernandoRevilla** · December 13th 2011, 08:59 AM

Originally Posted by maoro

#2 Assume $f(x)$ is continuous on $[a,b]$ , and for any $x\in[a,b]$ , there exists y\in[a,b] such that $|f(y)| \leq \frac{1}{2}|f(x)|$ . Prove that there exists a point $c \in [a,b]$ such that $f(c)=0.$

Hint: $g(x)=|h(x)|$ is continuous on $[a,b]$ .

**FernandoRevilla** · December 13th 2011, 09:06 AM

Originally Posted by maoro

I have just learned about Mean Value Theorem, is it similar to Lagrange Theorem?

Yes, it is another name.

btw, if $f(3)=4$ , we can just show that $f'(\xi)=1$ , but not $f(x)=x+1$

You can prove that $f'(x)=1$ for all $x\in (1,3)$ so $f(x)=x+C$ in $(1,3)$ etc ...

**maoro** · December 13th 2011, 09:09 AM

Originally Posted by FernandoRevilla

Yes, it is another name.

You can prove that $f'(x)=1$ for all $x\in (1,3)$ so $f(x)=x+C$ in $(1,3)$ etc ...

Is it because $f$ is differentiable, then if $f'(x)=1$ , we get $f(x)=x+C$ ?

**FernandoRevilla** · December 13th 2011, 09:22 AM

Originally Posted by maoro

Is it because $f$ is differentiable, then if $f'(x)=1$ , we get $f(x)=x+C$ ?

Right.

**maoro** · December 13th 2011, 09:25 AM

Originally Posted by FernandoRevilla

Hint: $g(x)=|h(x)|$ is continuous on $[a,b]$ .

What's the next step when $|f(x)|$ is continuous on $[a,b]$ ?
Can i simply state that there exists $c\in [a,b]$ such that

$|f(c)|\leq \frac{1}{2}|f(x_1)|\leq \frac{1}{2^2}|f(x_2)|\leq...\leq\frac{1}{2^n}|f(x_ n)|$

**FernandoRevilla** · December 13th 2011, 10:04 AM

Originally Posted by maoro

$|f(c)|\leq \frac{1}{2}|f(x_1)|\leq \frac{1}{2^2}|f(x_2)|\leq...\leq\frac{1}{2^n}|f(x_ n)|$

You have to prove that $c$ exists, so don't start with it. Construct a sequence $(y_n)$ in $[a,b]$ such that $0\leq g(y_n)\leq 2^{-n}g(a)$ , so $\lim_{n\to +\infty}g(y_n)=0$ etc.

**maoro** · December 13th 2011, 08:02 PM

I saw another problem which is similar to #1
Suppose that $f(x)=0$ and $f'(x) \leq 1$ . How large can $f(4)$ possibly be? (Done)
If $f(4) =4$ , then show that $f(x)=x$ for $0 \leq x \leq 4$

The question i want to ask is how can i prove $f(x)=x$ if i don't know $f$ is differentiable ?

**FernandoRevilla** · December 13th 2011, 10:06 PM

Originally Posted by maoro

The question i want to ask is how can i prove $f(x)=x$ if i don't know $f$ is differentiable ?

You can't, there are infinite functions satisfying $f(0)=0$ and $f(4)=4$ .

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