Respuesta :

Luv2Teach

Answer:

[tex]\frac{(x+5)(x-3)}{(x+5)(x+1)}[/tex]

Step-by-step explanation:

A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator.  It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x.  In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.

Using the continuity concept, it is found that the function with both a removable and a non-removable discontinuity is:

[tex]f(x) = \frac{(x - 1)(x - 2)}{(x - 1)(x - 3)}[/tex]

----------------------------------------------

Continuity:

A function is not continuous at points that are outside it's domain.

  • If the point outside the domain can be factored, it is removable.
  • If the point outside the domain cannot be factored, it is non-removable.

----------------------------------------------

An example can be taken considering a function as follows:

[tex]f(x) = \frac{(x - 1)(x - 2)}{(x - a)(x - b)}[/tex]

  • If a = 1, the term with [tex]x - 1[/tex] in the denominator can be factored with the term with [tex]x - 1[/tex] in the numerator, and this the function will have a removable discontinuity.
  • Now, if for example, b = 3, the term with [tex]x - 3[/tex] cannot be factored, and thus, the function will have a non-removable discontinuity.

The function is:

[tex]f(x) = \frac{(x - 1)(x - 2)}{(x - 1)(x - 3)}[/tex]

A similar example is given at https://brainly.com/question/23496517

Otras preguntas