1

enter image description here

I don't really need help with both parts because the proofs are going to be mostly quite similar so I'll just ask about part (a). So these are the things I know $$ f'(r)=\lim_{h->0} \frac{f(r+h)-f(r)}{h}\gt0. $$ I also see that the problem is trying to prove that if the derivative of a function at a specific point is greater than zero, than the original function is increasing sufficiently close to that point, in this case we want to be within $\delta$ of $r$. I also notice that the interval $(r, r+\delta)$ corresponds to a right sided limit, and the interval $(r-\delta, r)$ corresponds to a left sided limit. So somehow from all these facts I basically need to deduce the inequality $f(r) \lt f(x)$ for the first interval and $f(r) \gt f(x)$ for the second interval, but I'm not totally sure how.

This is a rough proof of the problem, but I don't know that I did it correctly Since f ’(r) exists and is > 0, the lim as h goes to zero of [f(r+h) – f(r)]/h > 0. This means the slope of the line through f(r+h) and f(r) is positive for values of h sufficiently close to zero as long as the range is bound by f ‘(r) – ε and f ‘(r) + ε. If we let |h| < δ and δ > 0, then r and r+h are contained within the interval (r – δ, r + δ). So since r + h, and r are both in these bounds and we have shown that the slope between f(r+h) and f(r) is positive for a value of h sufficiently close to r, then f(r) < f(x) when we pick a value of x to the right of r and yet to the left of r+δ, and also f(r) > f(x) when we pick a value of x to the left of r and yet to the right of r - δ

mmm
  • 1,879

2 Answers2

1

To do this think about what the derivative really means - that is, what is the limit, not our intuition, trying to say? In particular, the expression $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ says that the slope of the line from $f(x)$ to $f(x+h)$ can be constrained to within the range $(f'(x)-\varepsilon,f'(x)+\varepsilon)$ for any $\varepsilon$ by constraining $h$ to be close enough to $0$. Geometrically, this says that if we drew a line with slope a little more than $f'(x)$ and a line with slope a little less than $f'(x)$ through the point $(x,f(x))$, the function $f$ would, in some neighborhood of $x$, stay between the two lines.

The particular thing to note is that, suppose that $f'(x)>0$. Then, there must be some $\delta$ such that, if $|h|<\delta$, then $\frac{f(x+h)-f(x)}{h}>0$. This is from the definition of the limit - it represents bounding $f$ by a horizontal line (which has slope less than $f'(x)$) and some other line which, for now, we don't really care about. Of course, this implies that if $h$ is positive and within $(0,\delta)$, we get $$f(x+h)>f(x)$$and if $h$ is within $(-\delta,0)$, we get (since when we multiply the $h$ out of the denominator, we reverse the order as $h<0$): $$f(x+h)<f(x).$$

Milo Brandt
  • 61,938
  • From what you said, I tried to put together my own proof, so if you have the time, please see my edit and let me know if I've done something correctly or incorrectly. It would be very appreciated – mmm Nov 02 '14 at 18:10
  • 1
    It looks fine; you could explicitly state how you're using the limit: That is, you choose $\varepsilon$ to be something less than $f'(r)$ (so the lower bound is positive) and then are guaranteed a $\delta$ - it's just clearer to state the desired range then the "sufficiently close to $r$" condition. But, at some level, this is just a stylistic difference. – Milo Brandt Nov 02 '14 at 18:18
0

The fact that derivative exists means that for every $h >0$ there exists (we can choose it) $\varepsilon= x-x_0>0$ such that quotient

$$q=\frac{f(x_0+\varepsilon)-f(x_0)}{\varepsilon}$$

will satisfy

$f'(x_0)-h<q<f'(x_0)+h$ and $\forall x : (x_0, x)$ (Cauchy's definition of function's limit).

Therefore

$$f(x_0+\varepsilon)-f(x_0)>\varepsilon(f'(x_0)-h) ,\forall x : (x_0, x)$$

If we choose $h\leq f'(x_0)$ we are done.

4pie0
  • 481