Recent content by pmqable

I suppose you could differentiate both sides. Then you would get: f'(x)+f-1'(x)=2x f'(x)+1/(f'(f-1(x)))=2x but this equation still does not get rid of the f-1(x)... do you see something I don't?

by inverse i mean f(f-1(x))=x... e.g. ln(e^x)=x

so here's the question: if you have some equation relating a function, f(x), and its inverse, f-1(x), can you solve for the function? for example, solve for f(x): f(x)+f-1(x)=x^2 how about: f(x)+f-1(x)=g(x) my math teacher (AP calc) was stumped on this one... any thoughts?

thanks a lot lckurtz

Homework Statement dy/dx=y/x Homework Equations The Attempt at a Solution ok here's what I got... dy/dx=y/x so dy/y=dx/x. Then just integrate both sides and you get ln(y)=ln(x)+C. Next raise both sides to a power of e and you get y=e^(ln(x)+C). This can be rewritten y=e^ln(x)*e^C...

ok thank you very much... believe it or not, every integral i have ever done has resulted in a positive area. i just integrated x^2-4 from -1 to 1 and and sure enough, i got a negative area. lesson learned :)

oh wait it's negative haha... but it doesn't matter because -0=0 right?

positive?

sorry youre right the original equation should just be integral of ln(x)

the definite integral is: integral from 0 to 1 of ln(x). i used integration by parts (u=ln(x), du=1/x dx, dv=dx, v=x) to show that the integral is equal to: [x*ln(x)] (1,0) - integral from 0 to 1 of dx. this gives 1*ln(1)-0*ln(0)-(1-0) ln(1)=0, so the equation is now 0*ln(0)-1 0*ln(0) is an...

Homework Statement Here is the problem: ∫(1+sin(x))/(cos^2(x)) dx Also- how do you guys type equations in this? The quick symbols doesn't have a fraction bar or definite integral... Homework Equations sin^2(x)+cos^2(x)=1 The Attempt at a Solution I substituted 1-sin^2(x) for cos^2(x) and...

Homework Statement integral from 0 to pi of (x sinx)/(1+cos^2x)Homework Equations I just need to know if I did it rightThe Attempt at a Solution I got (5pi/6)-1/9

Yes I know it's a really stupid question, please don't make fun of me. But when they first defined this new function, the integral of 1/x, and had no graph or knowledge of how it looked, how could they know it was a logarithm? More specifically, the logarithm with base e.