Is the Function f(x) Differentiable at x=0 and x=1?

[destinc]

f(x)= x=1/x-1 if x ≤ 0
x^2-2x +1 if 0 < x < 1
ln x if x ≥ 1

Here's what I have so far
lim f(x)= lim x+1/x-1= -1, LHL= -1
x →0- x→0-
lim f(x)= lim x^2-2x+1= (1/2)^2 - [(2)1/2] +1= 1/4, RHL= 1/4
x→0+ x→0+
RHL ≠ LHL ∴ f(x) is not differentiable @ 0

Now I know I need to test x=1 for differentiability, but I am stuck on how to do that
I can take f ' (x)= 1/x which is continuous for all except 0, but not sure how to test limits
since there is no LHL with x ≥ 1

 

[SammyS]

f(x)= x=1/x-1 if x ≤ 0
x^2-2x +1 if 0 < x < 1
ln x if x ≥ 1

Here's what I have so far
lim f(x)= lim x+1/x-1= -1, LHL= -1
x →0- x→0-
lim f(x)= lim x^2-2x+1= (1/2)^2 - [(2)1/2] +1= 1/4, RHL= 1/4
x→0+ x→0+
RHL ≠ LHL ∴ f(x) is not differentiable @ 0

Now I know I need to test x=1 for differentiability, but I am stuck on how to do that
I can take f ' (x)= 1/x which is continuous for all except 0, but not sure how to test limits
since there is no LHL with x ≥ 1

Use of parentheses is needed for your expressions to say what I think you mean for them to say.

I think you mean:

f(x) =

(x+1)/(x-1) , if x ≤ 0
x²-2x +1 , if 0 < x < 1
ln x , if x ≥ 1​

Your limit for x→0⁺ is not correct. It's not 1/4.
\lim_{x\to 0^+}(x^2-2x +1) = 1

What is \displaystyle \lim_{\,x\to 1^-}(x^2-2x +1)\ \ ?

What is \displaystyle \lim_{\,x\to 1^+}\ln(x)\ \ ?

 

[destinc]

Thanks, you are correct about the setup of f(x).
I still get 1/4 when I plug 1/2 into x^2 - 2x + 1
1/2^2= 1/4, 2(1/2)=1
1/4 -1 + 1= 1/4
The limits set up for 1 helps, for some reason I thought I couldn't use the polynomial, but I finally found it in my notes.
Thank you

 

[destinc]

nvrmind previous post about RHL, I see my mistake.
So,
for the second part, I get
lim (x^2 - 2x +1)= 0
x→1-

lim lnx= 0
x→1+
Which means f(x0 is differentiable at 1 and final answer is
f(x) is differentiable for all real numbers x≠0
Correct?

 

[SammyS]

Yes, That is correct !