(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) =
This is probably ridiculously simple but I'm sleepy and I think I'm missing something... Hahaha
Do you know integration?
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.
antiderivative is another name for it
what is a function that has a derivative of 5x?
f'(x) - (f(b)-f(a)/b-a) is this a formula to find this answer? Hahaha
i dont think so, tht looks like the mean value thrm
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
Yeah that's MVT. I'm not seeing this in my notes, I did miss one class so it was probably covered then.
spose i asked you to find the derivative of kx^2 what would you try to do?
x^2+k*2x
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
Oh yeah my fault. I know derivatives just sleepy lol
now, we have 5x given as the derivative so we know it has to come from some kx^2 function right?
2kx = 5x when k=?
2.5
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?
Because it doesn't effect the derivative it just dissapears
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
so, let x=10 5(10)^2/2 + C = 0, what is C in this case?
-250
right, so f(2) = 5(2)^2/2 - 250
if f(-8) = 10 5(-8)^2/2 + C = 10, then C = 10-150-10 = -150 f(2) = 5(2)^2/2 - 150
Awesome! Makes sense thanks a lot. Not hard :-)
good luck ;)
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