Suppose that \(f\) satisfies the equation \[f(x + y) = f(x) + f(y) + x^2y + xy^2\] for all real numbers \(x\) and \(y\). Suppose further that \[\lim_{x→0} \frac{f(x)}x=1\] (a) Find \(f(0)\) (b) Find \(f'(0)\) (c) Find \(f'(x)\)
First off, if it's true for all real numbers, we know: f(x+0)=f(x)+f(0)+x^3*0+x*0^3 f(x)=f(x)+f(0) f(0)=0 Now for the derivative I am still distracted by the other question, so I'll come back to this but I imagine it just involves the chain rule or something.
cool, good on the first one :)
I'll give a hint if someone begs, but only one
f'(0) = lim x->0 { f(x) - f(0) } / (x - 0) = lim x->0 f(x) / x = 1
correct @ranga :)
Thanks.
I should have made it three posts so I could give a medal for each, but I'll only give it for the last one I guess
Yeah exactly, good answer. L'H definitely won't work here.
c) f'(x) = x^2 + 1
process please @ranga
I think I figured it out too late damn I'm writing furiously haha.
c) f'(x) = lim h-> 0 { f(x+h) - f(x) } / h = lim h-> 0 { f(x) + f(h) + x^2h + h^2x } / h = lim h-> 0 f(h) / h + x^2 + hx = 1 + x^2
\[f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow0}\frac{f(h)+x^2h+xh^2}{h}=\lim_{h\rightarrow0}\frac{f(h)}h+x^2+0=x^2+1\]
Yep.
lolz @ranga yay we have the same approach
Yay!!
very nice @ranga and @Kainui , I like questions that remind us that single-variable calc can take you a bit off-guard
Thanks @TuringTest
Where do you find these questions?
this is a very nice functional equation
^ that is a good question http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/ specifically this is the last question on the first exam at MIT http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/exam-1/session-22-materials-for-exam-1/MIT18_01SCF10_exam1.pdf
MIT OCW always throws the best equation,i had thought this questioin was a multivariable problem..not for single variable calculus
they sure do :) I'll post another shortly. not from the same source, of course
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