Arnav Posted June 7, 2021 Share Posted June 7, 2021 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? Quote Quote Quote Link to comment Share on other sites More sharing options...
studiot Posted June 7, 2021 Share Posted June 7, 2021 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. Link to comment Share on other sites More sharing options...
Country Boy Posted June 15, 2021 Share Posted June 15, 2021 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? Link to comment Share on other sites More sharing options...
Arnav Posted June 17, 2021 Author Share Posted June 17, 2021 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. Link to comment Share on other sites More sharing options...
Country Boy Posted June 18, 2021 Share Posted June 18, 2021 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. Link to comment Share on other sites More sharing options...
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