Is the Function f(x)= (x+1)/(1-√(1-x)) Derivable at x0=1⁻?

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In summary, the problem asks for the limit of a function when x gets close to 1 but is not equal to 1. The limit is found to be equal to the limit of the function when x gets close to 1 but is not equal to 1+2sqrt(1-x).
  • #1
mohlam12
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Hey everyone
I was going through some problems and I came across this one, it said to see if the fuction below is derivable in the point [tex]x_{0}=1^{-}[/tex]
[tex]f(x)=\frac{x+1}{1-\sqrt{1-x}}[/tex]

so I did this as usual,
limit of [tex]\frac{f(x)-f(1)}{x-1}[/tex] when x -> 1, and x<1 is equal to the limit of:

[tex]\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x}}[/tex]but I really don't know what to do after! I tried to multiply the top and bottom with "x+2sqrt(1-x)" but with no satisfying results.

Any help would be appreciated!
 
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  • #2
What do you mean by "derivable"?

Obviously, for [tex]x>1[/tex] you get complex solutions, and [tex]f(x)\rightarrow\pm\infty[/tex] as [tex]x\rightarrow0[/tex] (- from left, + from right)...

I wrote it as:

[tex]f(x)=\frac{(x+1)(1+\sqrt{1-x})}{x}[/tex]
 
  • #3
Maybe I meant differentiable... sorry!
what I need to find the limit of is this function [tex]f(x)=\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x})}[/tex] as [tex]x\rightarrow1^{-}[/tex]
Thank you
 
  • #4
Wait, can you copy exactly what the problem says please. I dunno, but f(1) is undefined, so the function is not differentiable at x = 1.
 
  • #5
VietDao29 said:
Wait, can you copy exactly what the problem says please. I dunno, but f(1) is undefined, so the function is not differentiable at x = 1.
Why undefined?

Because [tex]\sqrt{0}[/tex]?

As you approach from the left, it tends to 2 tho'...

(and the derivative goes to [tex]\infty[/tex]?)
 
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  • #6
mohlam12 said:
Maybe I meant differentiable... sorry!
what I need to find the limit of is this function [tex]f(x)=\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x})}[/tex] as [tex]x\rightarrow1^{-}[/tex]
Thank you

The limit of that expression is easy enough to find. Just substitute [tex]u = \sqrt{1-x}[/tex]

But I don't know what your question is asking, or, frankly, what you're trying to accomplish on the whole. Post the question verbatim, please.
 
  • #7
[tex]f(x)=\frac{x-1}{(x-1)(1-\sqrt{1-x})}+\frac{2}{1-x-\sqrt{1-x}}[/tex]
the first one's lim=1 and second one's lim=0
 
  • #8
J77 said:
Why undefined?

Because [tex]\sqrt{0}[/tex]?

As you approach from the left, it tends to 2 tho'...

(and the derivative goes to [tex]\infty[/tex]?)
Ack, ack, doing maths late is never good... :cry: :cry: :cry:
Okay, for mohlam12's question, have you considered factoring the [tex]\sqrt{1 - x}[/tex] out in the numerator?
[tex]\lim_{x \rightarrow 1 ^ -} \frac{x - 1 + 2 \sqrt{1 - x}}{(x - 1) (1 - \sqrt{1 - x})} = \lim_{x \rightarrow 1 ^ -} \frac{\sqrt{1 - x} (\sqrt{1 - x} - 2)}{(1 - x) (1 - \sqrt{1 - x})} = ...[/tex]
You can go from here, right? :)
Sorry for such confusion... My bad :frown:
 
  • #9
Yup, I can go from here, thanks. I just didn't have that idea to factor with [tex]\sqrt{1-x}[/tex]
 

1. What is the definition of a derivative?

A derivative is the rate of change of a function at a specific point. It represents how much the output of a function changes when the input is changed by a small amount.

2. How do you determine if a function is differentiable at a certain point?

A function is differentiable at a point if the limit of the difference quotient (Δx/Δy) exists as Δx approaches 0. In other words, the function must have a well-defined tangent line at that point.

3. Is the given function f(x)= (x+1)/(1-√(1-x)) differentiable at x0=1⁻?

From the definition of differentiability, we can see that the function is differentiable at x0=1⁻ if the limit of the difference quotient exists at that point. Therefore, we need to evaluate the limit as x approaches 1 from the left side. If the limit exists, then the function is differentiable at that point.

4. How do you calculate the derivative of a function?

The derivative of a function can be calculated using various methods such as the power rule, product rule, quotient rule, and chain rule. These rules involve taking the derivative of each term in the function and combining them according to the rule.

5. What does it mean if a function is not differentiable at a certain point?

If a function is not differentiable at a certain point, it means that the limit of the difference quotient does not exist at that point. This could be due to a sharp turn or corner in the graph of the function, or the function may not be defined at that point. In other words, the function does not have a well-defined tangent line at that point.

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