OpenStudy (anonymous):

Determine if the graph of f(x)=(x-1)/(x^2-1) is continuous for all real numbers, state "no discontinuities" and explain why using the conditions of continuity. If there are any discontinuities, please: Determine what x value the discontinuity occurs Determine if the discontinuity if removable or non-removable If removable, identify the value of the hole. If non-removable, show proof using the conditions of continuity.

7 years ago
OpenStudy (anonymous):

the / represents a fraction(:

7 years ago
OpenStudy (anonymous):

not continuous where denominator = 0 namely at -1 and 1

7 years ago
OpenStudy (anonymous):

now $\frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}=\frac{1}{x+1}$\]

7 years ago
OpenStudy (anonymous):

so the "discontinuity" at 1 is removable, because we just removed it. now you can plug in x = 1 if you like and get 1/2

7 years ago
OpenStudy (anonymous):

on the other hand the discontinuity at -1 is not removable, because it is still there. you still cannot replace x by -1 because the denominator will be 0

7 years ago
OpenStudy (anonymous):

so what would i write?

7 years ago
OpenStudy (anonymous):

the function is discontinuous at x = -1 and at x = 1 because the denominator is 0 at those two numbers. after doing the algebra we see that $f(x)=\frac{1}{x+1} \text{if} x\neq-1$

7 years ago
OpenStudy (anonymous):

therefore $lim_{x\rightarrow 1}f(x)=\frac{1}{2}$ and s the discontinuity at 1 is removable, since we can define the function there to be $f(1)=\frac{1}{2}$ making the function continuous at 1

7 years ago
OpenStudy (anonymous):

however at x=-1 the discontinuity is non removable because $lim_{x\rightarrow -1}f(x)$ does not exist

7 years ago
OpenStudy (anonymous):

alright, all of that?

7 years ago
OpenStudy (anonymous):

well they asked a bunch yes?

7 years ago
OpenStudy (anonymous):

If there are any discontinuities, please: Determine what x value the discontinuity occurs at x = 1 and at x = -1

7 years ago
OpenStudy (anonymous):

Determine if the discontinuity if removable or non-removable the discontinuity at 1 is removable by the reason stated above the discontinuity at -1is not removable by the reason stated above

7 years ago
OpenStudy (anonymous):

Yes(: Thanks!

7 years ago
OpenStudy (anonymous):

If removable, identify the value of the hole. 1 is removable and the value of the "hole" is 1/2 so we define the function there to be 1/2 and it is continous

7 years ago
OpenStudy (anonymous):

If non-removable, show proof using the conditions of continuity. x = -1 is not removable because the limit doesn't exist there, because the function has a vertical asymptote at x = -1 and the limit from one side is infinity and from the other is - infinity.

7 years ago