Math Help

[Viticius]

...Thanks to the ones that helped, We figured it out finally...Thanks tho...and for some reason my SIX guy wasn't working...So I winged it, but ya BEVERAGE + KEYBOARD = BAD, how is that for math..booya.

[Hachima]

That stuff is week 2, easy calc. If you need help with that basic of calc I suggest you go get some real tutoring help and learn how to do that stuff yourself. You sound like you want the answers and not help on solving them yourself. Maybe you should state how you are trying to solve them and what you don't understand exactly. Yeah, those are so basic I can do it in my head. Giving you the answers won't accomplish much though. Too many video games and not enough studying?

Edit:

When writing formulas with exponents use x^2 for x squared, not x2.

[Viticius]

Aint for me bro, this isn't my cup of tea, trying to help a friend. I don't care if this is week 32 calculus, I was just curious.

[Kuriin]

I probably did it wrong, so I deleted it.

But yeah, Hachima's right...learn how to write your math terms out correctly.

[Soriak]

Is that x^2? I don't mind looking at it, but no point in doing it with a wrong formula

[Kuriin]

When I was trying to do the problems, I couldn't figure out if it was ^2 or (2). I eventually did the ^2, but seeing as how I have yet to do Calculusssssss...oh well, at least I tried.

[Lyenae]

do you just want answers or you want to learn how to do it?

im white and nerdy so i'll help you out.

rate of change, slope, etc... is the derivative. simply take the derivative and evaluate for the mentioned x values. 1 and 8 use some useless theorem you won't use ever again that I forgot

1) C(x) = 0.05x^2 + 1606
(0.05(307)^2 + 1606 - 0.05(217)^2 -1606)/(307-217) = 26.2

2) f(x) = -4x^2
f'x = -8x
f'(3) = -8(3) = -24

3) C(x) = 0.06x^2 + 1521
C'(x) = 0.12x
C'(248) = 0.12(248) = 29.76

4) f(x) = 5-3x^2
f'(x) = -6x
f'(223) = -6(223) = 1338

5) N(t) = 12510 + 10t - t^2
N(5) = 12510 + 10(5) - 5^2 = 12535

6) f(x) = 7-5x-2x^2
f'(x) = -5 - 4x
f'(1) = -5 -4(1) = -9

7) N(t) = 1336 +11t - t^2
N'(t) = 11 -2t
N'(332) = 11 -2(332) = -55.4

8) f(x) = 2x - x^3
(2(7)-7^3 -2(0)-0^3)/(7-0) = -47

[Lyenae]

im assuming for 1 and 8 they want the difference in derivatives, but if someone understands it better they can fix my answers.

[Kuriin]

omfg I actually got some of them right when I did it. Yaaaaay.

[Ninajrr]

Mean Value Theorem, Lyenae, nyarr!

Should be f'(c) = [f(b)-f(a)] / [b-a]

1. 26.2
8. -47

The rest are all just first order derivatives of polynomials, tell your friend to look it up on Wikipedia, easiest type of derivative that there is. I imagine Lyenae got those correct, but I did not check.

[Hachima]

Originally Posted by Viticius

Aint for me bro, this isn't my cup of tea, trying to help a friend. I don't care if this is week 32 calculus, I was just curious.

I'd tell your friend the same thing then. If he is struggling at this point is screwed when he has to do stuff like using the chain rule to solve something.

If all you want is the answer and not actual help, well go to this site QuickMath Automatic Math Solutions

plug in the equation and solve it. Plug in the value for the result and you now have the slope.

[Soriak]

edit: nm, see solutions above

[Ninajrr]

Soriak, I don't think the first problem's first term is in parenthesis, nor the third, therefore the square only acts upon the x value, not the coefficient. Also, your math on your calculation of 1 is wrong, should be positive regardless.

[Soriak]

Oops, you're right about the parenthesis. But it definitely has to be negative, because the cost per produced item goes down: the 1606 are "fixed" costs that apply regardless of how much is produced, so as more is produced, every item becomes cheaper. (economics of scale)
(edit: unless it asked for total cost and not per piece, I suppose )


Excourse for the slope questions: You find the slope by differentiating an equation. Since no fractions, logarithms, roots, etc are part of the question, this couldn't be any easier:

You simply reduce the exponent of "x" by one and multiply the number before the "x" with the old exponent.

So:

5x^2 = (5*2)*x^(2-1) = 10x

f(x) = 10x^2
f'(x) = 20x
f''(x) = 20 (*x^0 = 1)
f'''(x) = 0

Any number without "x" simply disappears, since it's just a shift up and doesn't affect the actual graph. Obviously the same applies if the variable is something other than x, such as t:

f(t) = 10x + 10t^2
f'(t) = 20t

since "x" here is a constant and not a variable.

[Lyenae]

ty Ninjarr, i corrected my answers.