Hey everyone! I hope you all got to sleep, eat good food, and be with good people on your break.
I think having a running list of major concepts and proofs we have discussed over the semester may be helpful in our studying for the final exam. We can have the definition or theorem and then a list of the associated proofs or examples we have covered relating to the topic.
Also, this is a good forum to share and discuss the final problems sheet.
Made a single LaTeX document that we can all contribute to. At the moment it's organized by chapter and then by defs/theorems. Anyone feel free to restructure it however they like.
Oh, that's such a great idea! Will definitely contribute to this, thanks so much!
PS - I like how the saga of "Qurat was here" continues btw. ;$ $)
I could use some help/inspiration on #8 from the final problems sheet. The problem states:
What I am toying around with so far is using the MVT to say that there is some $c \in (0,1)$ such that $$f^{\prime}(c) = \frac{f(1) - f(0)}{1-0} = f(1) + 2 \geq 2$$
This shows $f(1) \geq0$.
From here I am stuck, if I am on the right track. I can see that $f^{\prime} \geq 2$ shows $f$ is increasing on [0,1]. Can I use this somehow along with $f(0) < 0 \leq f(1)$?
I have looked over both Corollary 5.3.3 & 5.3.4, but I am not seeing a connection.
Any help would be greatly appreciated!!!
Anyone have the list of terms we need to know for the final? @mark
What Mark down in the afternoon section was:
Here's the actual photograph of the definitions:
For number 3 on the final problems sheet, would this proof work? I don't really know exactly what it means by "use the difference quotient to prove". But I'm also stupid, so that might account for my confusion.
$3$. Suppose that $f$ and $g$ are defined on an open interval containing the point $c$ and that they are differentiable at $c$. Use the difference quotient to prove that $$(af + bg)'(c) = af'(c) + bg'(c)$$ for any real numbers $a$ and $b$.
Proof: Since $f$ and $g$ are differentiable at $c$, $(af + bg)'(c) = \lim_{y \to c} \dfrac{af(y) + bg(y) - (af(c) + bg(c))}{y-c} = \lim_{y \to c} \dfrac{af(y) - af(c)}{y-c} + \lim_{y \to c} \dfrac{bg(y) - bg(c)}{y-c} = a\left(\lim_{y \to c} \dfrac{f(y)-f(c)}{y-c}\right) + b\left( \lim_{y \to c} \dfrac{g(y) - g(c)}{y-c} \right) = af'(c) + bg'(c)$.
So... does this work or am I going to fail on Friday?
@gbrock Hey this looks good to me!! I think you meant $\frac{bg(y) - bg(c)}{y-c}$ instead of $\frac{bg(y) - by(c)}{y-c}$ on the second line at the beginning! b g (c) instead of by(c)
Thanks for catching that! I knew I probably messed up one letter here or there.
Suppose that $f$ and $g$ are real functions on the open interval $I$ that are both continuous at the point $c \in I$. Prove that $f + g$ is continuous at $c$.
Proof: Let $\epsilon > 0.$ Since $f$ and $g$ are continuous, then $$\exists \delta_1 > 0 \ni |x-c| < \delta_1 $$ implies $$|f(x) -f(c)| < \frac{\epsilon}{2} $$ and $$\exists\delta_2 > 0 \ni |x-c| < \delta_2 $$ implies $$|g(x) -g(c)| < \frac{\epsilon}{2}. $$
Choose, $\delta =$ min{$ \delta_1, \delta_2$} and when $|x-c| < \delta$ we have, $$ | (f(x) - g(x) )+ (f(c) - g(c))|$$
$$\leq | f(x) - f(c)| + | g(x) - g(c) |$$
$$ \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$
Therefore $f + g$ is also continuous at c.
Suppose $f: [a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and differentiable on (a,b) then there exists a point $c \in (a,b)$ where $ f'(c) = \frac{f(b) - f(a)}{b-a} $.
@lhouse1 Looks really good so far. I fixed a couple of little typos. A slightly more serious concern, though, is the choice of a single $\delta$ in terms of $\delta_1$ and $\delta_2$. Make sense?
@mark Ah yes! I think I fixed it, is that what you mean?
@lhouse1 Yes - that's it!!
I could use some help/inspiration on #8 from the final problems sheet. The problem states:
What I am toying around with so far is using the MVT to say that there is some $c \in (0,1)$ such that $$f^{\prime}(c) = \frac{f(1) - f(0)}{1-0} = f(1) + 2 \geq 2$$
This shows $f(1) \geq0$.
From here I am stuck, if I am on the right track. I can see that $f^{\prime} \geq 2$ shows $f$ is increasing on [0,1]. Can I use this somehow along with $f(0) < 0 \leq f(1)$?
I have looked over both Corollary 5.3.3 & 5.3.4, but I am not seeing a connection.
Any help would be greatly appreciated!!!
@cseagrav During the afternoon class, I realized that my lower bound on $f'$ needed to be larger. I think that $4$ should do it. That is, the problem should state that $f'(x)\geq4$ for all $x\in[0,1]$.
Sorry about that!!
thanks for the clarification.
I am having trouble with this one too. But are you sure it is necessary to use the MVT? I see your reasoning behind it I am just not sure if it is really needed. I will let you know if I get anywhere with it... @cseagrav
No i am not Positive but he lumped the last three problems into mvt problems if I remember correctly. I have not gotten anywhere with it either. Working a 10 hour day today so if you get any insight I would love that. Thanks.