Math Is Fun Forum

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-Anakin.

Yeah, I tried one a while back but it had trouble fitting and adjusting the shapes with more information.

If you recall, I hope you can post them sometime.

I'll start thinking of the content now.

Yeah, it is pretty terrible.

But I just wanted some of the ideas. Have you got some yourself?

For that one, if you memorize the multiplication table up to 12, it would really help.

The way that question would through my head is as so: I know 12x12=144. 132 is 12 less than 144 so the answer must be 11.

I guess I'd try to to figure out if I know any even simpler equation (12^2=144) and compare it to the current question (132/12).

But that's just me..

EDIT: Gosh, that was a bad bump. Didn't realize when the thread was made. Sorry.

Alright, will do!

Wow, that is REALLY cool. Quite simplified (at least for these basic functions).

Just to reconfirm to make sure I'm taking the right message home: For the derivative, the co-efficient is the power of the term multiplied by the original co-efficient, then all that is multiplied by the original variable to the power of n-1 (where n was the original power) to get the derivative.
Ex.
f(x)=4x^3-2x^2
f'(x) = 3 (4)(x)^2 - 2(2)(x)^1
f'(x) = 12x^2 - 4x
That is the correct concept behind what we discussed today, as to my understanding now. 

And, I looked at that link, and tried all the functions without looking at the answer, and got them all!

Hmm, the 2nd step is what I was concerned about. Thanks for clearing it up, that would make a bit easier to imagine. And the 3.125 was not even used, so no numbers.

And as for the last two formulas, I don't understand them. The D(x^n) = n * x^(n-1) : So for the first term we have -16(h+t)^2. Why isn't it -16(h+t)^1 then, since n-1 = 2-1 = 1.

Um. You said to observe where the maximum was. I did so and got 156.25.

Then I took f(t) = 100 t - 16 t^2 and subbed in 156.25 for f(t) to solve for t:

156.25 = 100 t - 16 t^2
16t^2-100t+156.25=0
16(t-3.125)^2 = 0
t=3.125

Or I could have used completing the square but that does not involve calculus so I didn't post it.

Or did you want me to use a specific method in mind? I thought about using the lim (h→0)⁡〖(f(x+h)-f(x))/h  〗formula but did not know where to go.

Very well. But what I meant was how could I find the maximum/minimum by using that algebraically?

Excellent! That can only apply for lines?

Hmm, and do you have some more thought-requiring questions?

I tried graphing them but the only differences I notice between y=5-x/15 and y=7-x/3 are the different y and x-intercepts. And the different slopes. Nothing else.

And for y=3x-5 and y=7-x/3, the derivatives of the two are negative reciprocals.

What am I missing?

Hurray!

And y=3x-5 has a slope or derivative of 3. The line y=7-x/3 has a slope or derivative of -1/3.