Question

For each of the following functions, determine if they're differentiable at x=0?
a) f(x)=sqrt(|x|)
b) f(x)=x*sqrt(|x|)
c) f(x)=sqrt(1+|x|)

Answer 1

Do you know about the left-derivative and the right-derivative?

Answer 2

um..no. Do you mean to derive, let's say for a, sqrt of -x and then sqrt of +x?

Answer 3

Oh wait, do you mean like limit as x approaches 0 from the negative side and then the limit as x approaches 0 from the positive side?

Answer 4

As with the other abs-function, you have to split up into two different functions, in this case one for x <=0 and one for x >0.
Each formula can be differentiated, so that is not the problem. The problem is the 0.
There should be the same value for f' whether you go from the right to 0 or from the left.

Answer 5

So for part a, the function is not differentiable because as x approaches 0 from the negative side, you can't sqrt a negative number.

Answer 6

Oh. I see.

Answer 7

No, because the |x| has been replaced with -x, and x is negative, so -x is positive...
Try yourself: (first function) f(-4) = sqrt(|-4|) =sqrt(4)=2 no sweat.

Answer 8

wait but for part a, the answer is that it is not differentiable.

Answer 9

Well, it isn't, but not because is it not defined. It is because the limit of f' as x approaches 0 is infinite, so there is a vertical tangent line.

Answer 10

oh...so to check if something is diff, I have to check if it's defined as well as the limit.

Answer 11

but then what about part b...

Answer 12

it's defined because when I plug in -1, it equals the same thing. But what about the limit?

Answer 13

Yes! And in the case there are two different formulas, the graphs of these must be connected smoothly to each other, otherwise you could not draw a tangent line.

Answer 14

b. f(x)=x*sqrt(|x|).

Answer 15

There are two different pieces of graph, but they are smoothly connected in x=0, so f is differentiable in 0.

Answer 16

but how do i prove this without actually drawing a graph using limits.

Answer 17

Left side: f(x)=x*sqrt(-x). Then\[f'(x)=1*\sqrt{-x}+x*\frac{ 1 }{ 2\sqrt{-x}}*-1=\sqrt{-x}+\frac{ -x }{ 2\sqrt{-x} }\]\[f'(x)=\sqrt{-x}+\frac{ 1 }{ 2 }\sqrt{-x}=\frac{ 3 }{ 2 }\sqrt{-x}\]When x approaches 0 from the left, -x goes to 0 (from the right) and the limit is 0.

Right side: everything the same, but w.o. the minus sign, limit of f' also 0, so left and right derivative equal, so f differentiable in 0.

Answer 18

Sorry, I have to go now... cyl

Answer 19

I'm back ;)

c. f(x)=sqrt(1+|x|)

Split up into two different formulas:

\[x >= 0 \rightarrow f(x)=\sqrt{1+x}\]\[x < 0 \rightarrow f(x)=\sqrt{1-x}\]We want to know if f is differentiable in x = 0.
For x >= 0 and for x < 0 there are these different derivatives of f:\[x >= 0 \rightarrow f'(x)=\frac{ 1 }{ 2\sqrt{1+x} }\]\[x<0 \rightarrow f'(x)=\frac{ 1 }{ 2\sqrt{1-x} }*-1=-\frac{ 1 }{ 2\sqrt{1-x} }\]f is differentiable in x = 0 if the left and right hand limit of f' are equal.

Calculate these limits to know what is going on in 0.