Find the x-values (if any) at which f is not continuous. If there are any discontinuities, determine whether they are removable. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.)
f(x) =
x
x^2 − 3
removable discontinuities x =
nonremovable discontinuities x =
To find the x-values at which ( f ) is not continuous, we need to look for points where the function is undefined or where there might be a jump or removable discontinuity.
The function ( f(x) = frac{x}{x^2 - 3} ) is a rational function. It will be undefined at values of ( x ) where the denominator equals zero, since division by zero is undefined.
For this function, the denominator ( x^2 - 3 ) equals zero when ( x = pm sqrt{3} ), because ( x^2 - 3 = 0 ) at those points.
Therefore, the nonremovable discontinuities for the function ( f(x) = frac{x}{x^2 - 3} ) are at ( x = -sqrt{3} ) and ( x = sqrt{3} ).
Since the function is continuous everywhere else (except at these points), there are no removable discontinuities for this function.
Therefore, the answers are:
Removable discontinuities: DNE
Nonremovable discontinuities: ( x = -sqrt{3}, sqrt{3} )