# Differentiable

• Apr 22nd 2007, 12:10 PM
learn18
Differentiable
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
• Apr 22nd 2007, 12:15 PM
Jhevon
Quote:

Originally Posted by learn18
a) f(x) = { 3x-2 if x<1 and x^3 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
• Apr 22nd 2007, 12:17 PM
CaptainBlack
Quote:

Originally Posted by learn18
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

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
• Apr 22nd 2007, 12:18 PM
Jhevon
Quote:

Originally Posted by learn18
Determine if each function is differentiable at x=1. If it is find the dirivative, if not explain y not.
b) f(x) = { 2x+1 if x<1 and x^2 if x>=1

See the graph below. Obviously f(x) is not continuous at the desired point, so we cannot find the derivative. You can show this by showing that the left and right hand limits are different (do you know how to do that?)

EDIT: Nevermind, CaptainBlack did it for you
• Apr 22nd 2007, 12:21 PM
Jhevon
Quote:

Originally Posted by CaptainBlack
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