Did you mean to let epsilon = f'(c)/2?
Because that would solve it.
It would then be that f'(c) - f'(c)/2 < f'(x) => f(c)/2 < f'(x). Which means that f'(x) is strictly greater than 0. Thus f(x) is strictly increasing.
Is this right?
Thanks for the reply. How does taking epsilon = f(c)/2 get me that f'(x) > 0?
I plugged it in and I end up getting that f'(c) - f(c)/2 < f(x) < f'(c) + f(c)/2
I understand, though, that finding f'(x) > 0 solves it. I'm just curious about this middle step. I assume I have to show that f'(c)...
Homework Statement
Suppose f is differentiable on J, c is in J0 and f'(c) > 0. Show that if f' is continuous at c, then f is strictly increasing on some neighborhood of c
Homework Equations
Strictly increasing: If x < y then f(x) < f(y)
Continuous: For all epsilon > 0 there exists a...
Homework Statement
Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.
Let f be continuous at x=c and f(c) > 1
Show that there exists an r > 0 such that \forall x \in B(c,r) \bigcap D : f(x) > 1
Homework Equations
\forall \epsilon >...
ah, duh, thanks. 3 < |x+2| < 5 (since x+2 is positive with or without the absolute value)
So then
\frac{|x-2||x+2|}{4} \leq \frac{5*|x-2|}{4} < \epsilon
Thus |x-2| < \frac{4}{5}\epsilon
Thus if \delta = \frac{4}{5}\epsilon \Rightarrow |f(x) - \frac{1}{4}| < \epsilonThanks so much.
Yeah I got that far. I know that
|(x-2)||(x+2)| < 4\epsilon
But that still leaves that (x+2) there (which I can remove the absolute value from since it has to be positive). I can't have delta dependent on x.
I'm probably missing something blatant and obvious right here.
Homework Statement
Use the definition of limits to show that
lim _{x\rightarrow 2} \frac{1}{x^{2}} = \frac{1}{4}
Homework Equations
\forall \epsilon > 0, \exists \delta > 0, x \in D, 0 < |x-2| < \delta \Rightarrow |\frac{1}{x^{2}} - \frac{1}{4}| < \epsilon
The Attempt at a...
A guy on the math forums wrote out the solution for me and it's obviously simple now that I see it:
\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {\left( {x + c} \right) - \left( {a + c} \right)} \right| < \delta \, \Rightarrow \left| {f(x + c) - f(a + c)} \right|\, < \varepsilon...
I think you came in while I was fixing the <tex> stuff actually.
Exactly how do you know that f(x) = g(x+c)? I don't remember a lot of my early calculus tricks. And why would that help?
Suppose f:D\rightarrow \Re, c \in \Re and g(x) = f(x-c)
1) What's the Domain of g?
I think it's \Re, am I right?
2) Suppose that f is continuous at a \in D \Leftrightarrow g is continuous at c + a
So far I have this:
(\Rightarrow) Assume f is continuous. Then:
\forall \epsilon...
Why couldn't this book have just said that? That makes complete sense. My book doesn't like making sense. I don't think this was written to be a textbook at all, because it certainly doesn't read like it's trying to teach!
Ok so for the question I'm looking for points a,b,c, etc. where I can...
This is the definition of accumulation point that my book gives:
A is an accumulation point of S \subset \mathbb{R}, \forall \epsilon > 0, S \bigcap B(A;\epsilon) is infinite.
The book I have gives horrible examples on what accumulation points actually are (contradicting itself two out of...
So does this seem sufficient?:
A relation R on A has 16 possible ordered pairs. Let R be a relation on A with 15 ordered pairs excluding aRc. Since all the other remaining pairs are in R, then aRb and bRc. However, since a does not relate to c, R is not transitive.
Ok, I see the problem at least. I couldn't visualize with symmetric originally. But if you have (4,3) and (3,4) but not (4,4) you aren't transitive. Well at least that group isn't, but it's supposed to be for any a,b,c right?
Now the problem is... how do I put that in words?