wyantjs Wrote: ------------------------------------------------------- > Can it just be something like a continuous > piecewise function? A little more complex.
alright, I give up. I have tried inverse functions, trig functions, inverse trigs, hyperbolics, all types of logs, etc…and cannot come up with this. This is sad considering I have a masters in math.
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maratikus Wrote: ------------------------------------------------------- > > > Construct a function f(x) defined for all real x > such that it has derivative in x = 0 and x = 1 and > has no derivative anywhere else. f(x) = {x^2, if x is rational; -x^2, if x is irrational} is only differentiable at x=0 where f’(0)=0, it is discontinuous (and hence not differentiable) anywhere else shift that function horizontally by 1 unit to the right -> g(x)=f(x-1), now g(x) is only differentiable at x=1 and it is discontinuous everywhere else. now combine the two functions: h(x) = {f(x), if x<1/2; g(x), if x>1/2}. i think that should be differentiable at x=0 and x=1 and discontinuous everywhere else?
why is that discontinuous everywhere else? we have: df/dx = {2x if x is rational; -2x if irrational}. This is defined for all x. correct me if I am missing something.
in order for the derivative at x to exist, [f(x+h)-f(x)]/h must exist in the limit h->0. that essentially means the numerator must also tend to zero, so f(x+h) must converge to f(x) as h->0. this isn’t the case here because the way the function is defined, it jumps from positive to negative when the argument x switches from rational to irrational (there is a rational between any two irrationals, and there is an irrational between any two rationals, i.e. both rationals and irrationals are dense on the real line). the only limit that exists is when x=0
Nice solution, Mobius! I was originally given that problem on the exam my freshman year in college. The way I constructed the function was multiplying Dirichlet function D(x) //which is equal 1 for rational x and 0 for irrational x// by x^2*(x-1)^2. You pretty much multiplied x^2 by 2*D(x)-1 and (x-1)^2 by 2*D(x)-1 and then lined those two functions … same idea.
Mobius Striptease Wrote: ------------------------------------------------------- > in order for the derivative at x to exist, /h must > exist in the limit h->0. that essentially means > the numerator must also tend to zero, so f(x+h) > must converge to f(x) as h->0. this isn’t the case > here because the way the function is defined, it > jumps from positive to negative when the argument > x switches from rational to irrational (there is a > rational between any two irrationals, and there is > an irrational between any two rationals, i.e. both > rationals and irrationals are dense on the real > line). the only limit that exists is when x=0 ahhhh…gotcha. nice solution indeed.
Man, I wish I was better at math …
This isn’t a very hard calculus question. Problem is that i dont remember squat from my math classes.
storko Wrote: ------------------------------------------------------- > Man, I wish I was better at math … Does it mean you are good at other things?