Determine if each function is differentiable at x=1. If it is find the dirivative, if not explain y not.
a) f(x) = { 3x-2 if x<1 and x^3 if x >=1
b) f(x) = { 2x+1 if x<1 and x^2 if x>=1
We are differentiable as long as we are continuous. so we know f(x) is differentiable at x = 1 if it is continuous at x = 1. you can show this by finding the left and right hand limits. i will just show you the graph, so you can see it's continuos.
as you see, f(x) is differentiable here.
using f(x) = x^3
=> f ' (x) = 3x^2
Because these functions are both piecewise differentiable all we need to
show is that:
lim_{x->1-} f(x) = lim_{x->1+} f(x),
and that:
lim_{x->1-} f'(x) = lim_{x->1+} f'(x),
If both of these hold then the function is differentiable at x=1, otherwise they
are not.
For a) lim_{x->1-} f(x) = 1, lim_{x->1+} f(x) = 1, and lim_{x->1-} f'(x) = 3,
lim_{x->1+} f'(x) = 3, so f is differentiable at x=1.
For b) lim_{x->1-} f(x) = 3, lim_{x->1+} f(x) = 1, so f is not differentiable
at x=1.
RonL
I was not aware that we have to check the left and right hand limits for f', i thought once it checked out for f we can assume it would check out for f'. is this an incorrect assumption? can you give a counter example, where f is continous at the point but f' isn't
EDIT: nevermind, learn18 provided us with a counter example in a subsequent problem, see http://www.mathhelpforum.com/math-he...erntiable.html