I was told to find the range of a function f(x) = (x²-1)/(x-1) where x cannot be 1.

I know the proper solution is this : 

Since x cannot be 1, f(x) reduces to x+1 therefore Range = R - {2}

Bu at my first attempt I did this:

let y = (x²-1)/(x-1)

     yx - y = x²-1

    x² - yx + y -1 =0

For x to be real, the discriminant of this equation should be >= 0

Therefore y² - 4(y-1) >= 0

(y-2)² >= 0

y can take all real values.

Now, where did I go wrong? Why is 2 also coming in the range with the second method?

How do I distinguish when to find the range using the 2nd method and when not?

 

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1 hour ago, Arnav said:

x² - yx + y -1 =0

This leads to

put y = 2 then

x² -2x +1 =0

(x-1) (x-1) = 0

x =1.

On 6/7/2021 at 1:48 PM, Arnav said:

I was told to find the range of a function f(x) = (x²-1)/(x-1) where x cannot be 1.

I know the proper solution is this : 

Since x cannot be 1, f(x) reduces to x+1 therefore Range = R - {2}

Bu at my first attempt I did this:

let y = (x²-1)/(x-1)

     yx - y = x²-1[/quote]

When you did this, you removed the discontinuity.

    x² - yx + y -1 =0

For x to be real, the discriminant of this equation should be >= 0

Therefore y² - 4(y-1) >= 0

(y-2)² >= 0

y can take all real values.

Now, where did I go wrong? Why is 2 also coming in the range with the second method?

How do I distinguish when to find the range using the 2nd method and when not?

 

 

On 6/15/2021 at 5:03 PM, Country Boy said:

On 6/15/2021 at 5:03 PM, Country Boy said:

When you did this, you removed the discontinuity.

 

could you please elaborate? I have not studied about continuity and discontinuity yet.

On 6/17/2021 at 1:36 PM, Arnav said:

could you please elaborate? I have not studied about continuity and discontinuity yet.

A function, f(x), is continuous at x= a if and only if

1) f(a) exists

2) $\lim_{x\to a} f(x)$ exists

3)$\lim_{x\to a} f(x)= f(a)

Since (3) requires both (1) and (2) often we just state (3).

Of course, if a function is NOT continuous at x= a, it is discontinuous there,  

 

Here, [tex]f(x)= \frac{x^2- 1}{x- 1}[/tex] is not continuous at x=1 becaue $f(1)= \frac{0}{0}$ does not exist.  As long as x is NOT 1, [tex]\frac{x^2- 1}{x- 1}= x+ 1[/tex].  g(x)= x+ 1 exists at x= 1, g(1)= 2 and $\lim_{x\to 1} x+ 1= (\lim_{x\to 1} x)+ 1= 1+ 1= 2$ so this new function, g(x)= x+ 1, is continuous at x= 1.