Comrade, Thank you very much for your thoughtful message! Great to hear from you. This year has certainly been an adjustment! Although the environment is not nearly as exciting, motivating or entertaining as that which I'm used to, I believe I made the right choice...a necessary choice. It would be impossible to do what I'm currently achieving while simultaneously carrying out the responsibilities of a classroom teacher. Much of the work I've been doing, especially the interactive digital resource development, has been inspired by your class. Your engagement and feedback over our year and a half together have been pivotal in shaping the direction of my current initiatives. Although I'd definitely like to return to the classroom in the future, I have years of work ahead of me that I hope to first complete in this role. I do think of your class often. In fact, just yesterday I was telling a friend about our class's brief obsession with jigsaw puzzles and the ultimate disappointment of the missing piece from the 3000-piece monster. I'm still convinced that someone out there has that piece and will one day mail it to me! So many great memories from our days in 272! I really hope to hear from you all over the years to come. If you're ever in the area, don't hesitate to stop by and say hi. I'm often at the board office on Fairview. Send me a message ahead of time and I'll be sure to get you security clearance...haha! Regarding the website, I've had to temporarily make it available only through member login (your credentials will work as long as you remain a "site member"). As a result, it no longer shows up in browser searches. Nevertheless, this forum and all the other content will remain available. The outlines and resources are actually still being used by other SJC classes. Once again, thanks for your kind words. I sincerely hope that you'll keep in touch. Take care and best wishes for your second term! G

This makes me very happy and very frightened at the same time!

Haha! I love it! I'll definitely miss you all as well, but I'm certain I'll see you often at SJC. I hope you're having an awesome summer!

Yes, if needed.

Not that I can remember. I'll search for it tomorrow in class.

Nope!

The original limit is indeterminite form 1^infinity. To help evaluate the limit of f(x) we investigate the limit of lnf(x). This allows us to "bring down" the exponent using a log law. The limit of lnf(x) is then easy to find. We weren't asked to find the limit of lnf(x) though...we were asked to find the limit of f(x). So, we sneakily write f(x) as e^lnf(x). We're using a log law here....a^loga[x]=x (the a is the log base). Remember, ln is just a log with base e. So, the limit of f(x) is the same as the limit of e^lnf(x), but with the latter representation, we can use our knowledge that the limit of lnf(x) equals 1. Make sense?

You just need to know the sin, cos and tan derivatives, and perhaps their reciprocals (csc, sec and cot) if you don't want to come up with them on the spot. Don't worry about the inverse trig derivatives for this exam.

I am.

Me thinks it's a bit easier to attack this one with the subtraction formula instead of the double angle formula. Check it out...

Does this help? By the way, you missed a negative in the second fraction's denominator.

Out all day today for Mother's Day stuff. I'll try to post an answer later, but if I can't, stop by my room sometime tomorrow and we'll go over this.

They were given out in class.

Hi! Unfortunately, I'll be away from my computer for the rest of the evening. Come see me tomorrow and we'll get your questions answered. Shouldn't take long!

Did you bake this? Looks delicious!

For part (a), they want you to look at the graphs and state where (x+5)/(x-1) is less than 4. Since y=(x+5)/(x-1) is the green graph and y=4 is the blue graph, you need to state the x-values for which the green graph is below the blue graph (ignore the red graph for now). This occurs from -∞ to the vertical asymptote at x=1 and from 3 to ∞ . I've highlighted these regions in the picture below. Therefore the answer is (-∞,1) and (3, ∞ ). For part (b), do the same thing, but this time state where the red graph is above the green graph. Looks like its roughly (-0.5, 1) and (2, ∞). See picture below. This question would be a lot easier if there were only two graphs and not a third one getting in the way!

Move everything to the left side of the inequality and combine the terms using a common denominator of sinx. Create a chart using the zeros and vertical asymptotes. You'll need to remember a bit about solving trig equations. We can solve the problem in full tomorrow in class!

and messy!