# Monotonically Increasing Proof

• Jan 16th 2011, 08:52 AM
mathematic
Monotonically Increasing Proof
Suppose the function
f has the following four properties:
1.
f is continuous for x >=0;

2.
f'(x) exists for x > 0;

3.
f(0) = 0;

4.
f'is monotonically increasing.
Put

g
(x) =f(x)/x

;
(x > 0)
and prove that
g is monotonically increasing. To better understand the various hypotheses,

sketch the graph of a couple of typical functions
f and interpret g geometrically.

I started by trying some various functions for f.
I tried f(x)=x,f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch.
Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the othr, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?
• Jan 16th 2011, 10:23 AM
Opalg
Quote:

Originally Posted by mathematic
Suppose the function f has the following four properties:
1. f is continuous for x >=0;
2. f'(x) exists for x > 0;
3. f(0) = 0;
4. f' is monotonically increasing.
Put g(x) = f(x)/x; (x > 0)
and prove that g is monotonically increasing. To better understand the various hypotheses, sketch the graph of a couple of typical functions f and interpret g geometrically.

I started by trying some various functions for f. I tried f(x)=x, f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch

Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the other, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?

Condition 4. says that $\displaystyle f'$ (the derivative of $\displaystyle f$) must be monotonically increasing. That condition is not satisfied by either of your two trial functions. So yes, you should have used different functions for $\displaystyle f$.
• Jan 16th 2011, 10:28 AM
mathematic
Would f(x)=x^2 and f(x)=x^3 work? Both of those are increasing proved x>=0.
Now f'(x)=2x and f'(x)=3x^2. Increasing when x>0
• Jan 16th 2011, 10:29 AM
Drexel28
Quote:

Originally Posted by mathematic
Suppose the function
f has the following four properties:
1.
f is continuous for x >=0;

2.
f'(x) exists for x > 0;

3.
f(0) = 0;

4.
f'is monotonically increasing.
Put

g
(x) =f(x)/x

;
(x > 0)
and prove that
g is monotonically increasing. To better understand the various hypotheses,

sketch the graph of a couple of typical functions
f and interpret g geometrically.

I started by trying some various functions for f.
I tried f(x)=x,f(x)=x^1/2. Those are the two I looked at so far. I was trying to see if maybe there were better ones to sketch.
Now interpreting g geometrically has me a little confused. I guess that would mean just sketching g. Now if f(x)=x, then g is the graph of 1. That doesn't seem right for the g to be monotonically increasing. For the othr, we have g=x^1/2/x=x^-1/2. Again, I get the feeling that g is not monotonically increasing. So, maybe I should have used different functions for f?

Merely note that if one let's $\displaystyle \displaystyle g(x)=\frac{f(x)}{x}$ then one has $\displaystyle \displaystyle g'(x)=\frac{f'(x)x-f(x)}{x^2}$ so it suffices to prove that $\displaystyle f'(x)x\geqslant f(x)$ for all $\displaystyle x>0$. To see this let $\displaystyle x_0>0$ be fixed. Then, by assumption we may apply the MVT on $\displaystyle [0,x_0]$ to get that $\displaystyle \displaystyle \frac{f(x_0)}{x_0}=\frac{f(x_0)-f(0)}{x_.-0}=f'(\xi)$ for some $\displaystyle \xi\in(0,x_0)$ but since $\displaystyle f'$ is increasing we may conclude that $\displaystyle \displaystyle f'(x_0)\geqslant f'(\xi)=\frac{f(x_0)}{x_0}$. Since $\displaystyle x_0$ was arbitrary the conclusion follows.
• Jan 16th 2011, 10:45 AM
mathematic
Well, I'm supposed to show this without the MVT. That theorem doesn't come till the next section in the book.
• Jan 16th 2011, 10:49 AM
Drexel28
Quote:

Originally Posted by mathematic
Well, I'm supposed to show this without the MVT. That theorem doesn't come till the next section in the book.

Ok, what machinery do you have?
• Jan 16th 2011, 10:58 AM
mathematic
Ok, the chapter I'm working out of is on the derivative. In particular, the last section we did was on derivatives and the intermediate value property. I have the limit definition of a derivative, definition of differentiable, combinations of differentiable functions(product rule, quotient rule, etc), chain rule, Darboux's Theorem, Interior Extremum Theorem.
Interior Extremum Theorem says:Let f be differentiable on (a,b). If f attains a maximum value at some point c in (a,b) then f'(c)=0.
Darbooux's Theorem says:If f is differentiable on an interval [a,b] and if d satisfies f'(a)<d<f'(b) then there exists a point c in (a,b) where f'(c)=d.

Previous chapters have been on sequences, series, open/closed/compact sets, limits and continuity.
• Jan 16th 2011, 03:12 PM
mathematic
Does any of that work? If none of what I have so far will be helpful, I will use the MVT, but I was iffy on using that since we only talked about that in the last 5 min of class.
• Jan 16th 2011, 04:20 PM
LoblawsLawBlog
Are you sure this comes before MVT? This exact problem is in Baby Rudin and it comes after MVT there. But if you really can't use MVT then I can't help you because my solution is the same as Drexel's
• Jan 16th 2011, 05:00 PM
mathematic
Ok well I could use the mvt. This wasn't assigned from our book, but all other problems assigned were out of the previous section, so I just assumed that this one I would use the previous section.