hello, need your help
find all function differentiable over
is differentiable over
Ok not to be a pain but the word is differentiable not derivable.
Taking the square root
f ' ^2 -f = 1 or -1
Differentiate -- in either case
2f " f ' -f ' = 0
f ' (2f"-1) = 0
Now solve f ' = 0 (use original equation to find values of c) and 2f " - 1 = 0
That should be enough to get you to the end
Sorry I read it too quickly I was seeing
[f '^2 -f]^2 = 1 for the original
For (f ')^2 - f^2 =1
the process is still the same
2f ' f "- 2 f f '= 0
f ' = 0 or f '' -f = 0
yields f = c from which there are no solutions
or f " -f = 0 which yield Acos(x) + Bsin(x) now apply f '(0) = 1
and finish
Ok I've made every possible mistake on this to now--thats what i get for hurrying
we get f = Ae^t + Be^-t
f ' (0) yields A+B =1
Substituting into original equation yields -4AB =1
Now solve --solutions are when A = (2^1/2+1)/2 or (1-2^1/2)/2
with corresponding values for B = 1 -A
Sorry for the mistakes
i'm sure you can find more mistakes to make if you tried
it happens to the best of us, don't worry about it. mistakes were corrected before they caused much damage.Sorry for the mistakes
did you double check your solutions for A?
i'm not going to check them, i trust you
I rechecked for both A and B in the original both by hand and with Mathcad to make sure-- thanx for the kind words
But still I always told my students to slow down --Calculus is more about patience than intelligence I'd say--too bad I don't follow my own advice.
That's what happens a year removed from teaching--skills are slipping