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Mathematics 8 Online
OpenStudy (anonymous):

(1 pt) Suppose that f'(x) = 5 x for all x. Compute f(2) if: f(10) = 0 f(2) = f(-8) = 10 f(2) =

OpenStudy (anonymous):

This is probably ridiculously simple but I'm sleepy and I think I'm missing something... Hahaha

Parth (parthkohli):

Do you know integration?

OpenStudy (anonymous):

I actually don't think we've covered integration no, is there another way to do this? Or if integration is a short topic to learn maybe it's hidden in my notes somewhere.

OpenStudy (amistre64):

antiderivative is another name for it

OpenStudy (amistre64):

what is a function that has a derivative of 5x?

OpenStudy (anonymous):

f'(x) - (f(b)-f(a)/b-a) is this a formula to find this answer? Hahaha

OpenStudy (amistre64):

i dont think so, tht looks like the mean value thrm

OpenStudy (amistre64):

derivatives are a basic process that finds one function from another. if we are given a function, f(x), we can determine its derivative f'(x) real simple by following a few simple rules. when we want to work this in reverse, its tricky but not as tricky as youd think ... we simply have to undo the rules

OpenStudy (anonymous):

Yeah that's MVT. I'm not seeing this in my notes, I did miss one class so it was probably covered then.

OpenStudy (amistre64):

spose i asked you to find the derivative of kx^2 what would you try to do?

OpenStudy (anonymous):

x^2+k*2x

OpenStudy (amistre64):

well, assuming you just tried a product rule: k is an arbitrary constant and the derivative of a constant is 0 so 0x^2 + 2kx = 2k x

OpenStudy (anonymous):

Oh yeah my fault. I know derivatives just sleepy lol

OpenStudy (amistre64):

now, we have 5x given as the derivative so we know it has to come from some kx^2 function right?

OpenStudy (amistre64):

2kx = 5x when k=?

OpenStudy (anonymous):

2.5

OpenStudy (amistre64):

5/2 yes :) now, here is the trickiness of it all. there are an infinite amount of functions that share the same derivative and they all differ by a constant 5/2 x^2 + C defines them all, why?

OpenStudy (anonymous):

Because it doesn't effect the derivative it just dissapears

OpenStudy (amistre64):

fair enough so, the question is asking you to really determine a value for C that makes f(10)=0 or when f(-8)=10

OpenStudy (amistre64):

so, let x=10 5(10)^2/2 + C = 0, what is C in this case?

OpenStudy (anonymous):

-250

OpenStudy (amistre64):

right, so f(2) = 5(2)^2/2 - 250

OpenStudy (amistre64):

if f(-8) = 10 5(-8)^2/2 + C = 10, then C = 10-150-10 = -150 f(2) = 5(2)^2/2 - 150

OpenStudy (anonymous):

Awesome! Makes sense thanks a lot. Not hard :-)

OpenStudy (amistre64):

good luck ;)

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