Using derivative with apparent increasing values

[jfy4]

Homework Statement

"If $f'(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$ show that there is a $\delta>0$ so that

$$f(x)<f(x_0)<f(y)$$

whenever $x_0-\delta<x<x_0<y<x_0+\delta$. Does this assert that $f$ is increasing in the interval $(x_0-\delta,x_0+\delta)$?"

Homework Equations

$$f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

The Attempt at a Solution

To be honest, the only way I could get this to work is by assuming the function is increasing given that the derivative is greater than zero, but it appears that is something I am suppose to show in the first place. The direction I want to take in my mind is that I can make two chords on the function, one from $f(x)$ to $f(x_0)$ and the other from $f(x_0)$ to $f(y)$ using the given data that the derivative is greater than zero. But I'm not sure how to lump in the epsilon-delta stuff.

 

[Mark44]

Homework Statement

"If $f'(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$ show that there is a $\delta>0$ so that

$$f(x)<f(x_0)<f(y)$$

whenever $x_0-\delta<x<x_0<y<x_0+\delta$. Does this assert that $f$ is increasing in the interval $(x_0-\delta,x_0+\delta)$?"

Homework Equations

$$f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

The Attempt at a Solution

To be honest, the only way I could get this to work is by assuming the function is increasing given that the derivative is greater than zero, but it appears that is something I am suppose to show in the first place.

Right, you need to show this, so you can't just assume it.

The direction I want to take in my mind is that I can make two chords on the function, one from $f(x)$ to $f(x_0)$ and the other from $f(x_0)$ to $f(y)$ using the given data that the derivative is greater than zero. But I'm not sure how to lump in the epsilon-delta stuff.

Start with this,
$$f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}$$

but write it in terms of the epsilon-delta definition of a limit.

 

[jfy4]

Here is my attempt at the solution after some more time... :

First let

$$g(x)=\frac{f(x)-f(x_0)}{x-x_0}$$

then we know that

$$\lim_{x\rightarrow x_0}g(x)>0.$$

Then pick an $\epsilon=g(x_0)/2>0$, then there exists a $\delta>0$ such that for all $x\in(x_0-\delta,x_0+\delta)$, $g(x)\in(g(x_0)-\epsilon,g(x_0)+\epsilon)$. Then pick an arbitrary element in the interval and note that if $x<x_0$, then $f(x)<f(x_0)$, and if $y>x_0$, then $f(y)>f(x_0)$, as demanded by the strict positive inequality. Then there exists a $\delta$ such that if $x_0-\delta<x<x_0<y<x_0+\delta$, then $f(x)<f(x_0)<f(y)$. $\blacksquare$