Respuesta :
Answer:
There is a vertical asymptote at x = 0.
There is a removable discontinuity at x = –a
Step-by-step explanation:
The given rational function is [tex]f(x)=\frac{(x+a)(x+b)}{x^2+ax}[/tex]
We can rewrite the function by factoring the denominator as
[tex]f(x)=\frac{(x+a)(x+b)}{x(x+a)}[/tex]
Vertical asymptotes: Vertical asymptotes occurs where the denominator is zero.
Removal discontinuity: If we have some common factor in the ration function which can be canceled out then at that point, there would be a removal discontinuity. This is same as a hole in the graph.
In the given rational function, we have (x+a) is common in both numerator and denominator. Thus, the removal discontinuity is given by
x+a =0
x = -a
For vertical asymtotes:
x = 0
Therefore, 2 and 5 are correct options.
The asymptote and discontinuity of the function, by its analysis is found to be as
- There is a vertical asymptote at x = 0 for the given function.
- There are no removable discontinuity for the given function.
When do we get vertical asymptote for a function?
Suppose that we have the function f(x) such that it is continuous for all values < a or > a and have got the values of f(x) going to [tex]\infty[/tex] or [tex]-\infty[/tex] as x goes near a , and being not defined at x = a, then at that point, there can be constructed a vertical line x = a and it will be called as vertical asymptote for f(x) at x = a
What are removable discontinuities?
If a function f(x) has
[tex]lim_{x \rightarrow a^-}f(x) = L = lim_{x \rightarrow a^+}f(x)[/tex]
but these limits doesn't equate to the value of the function f(a).
(Also, note that we need L, and f(a) both finite defined numbers)
then this type of discontinuity is called removable discontinuity and we can remove it by defining that
[tex]f(a) = L[/tex]
The given function is
[tex]f(x)=\dfrac{(x+a)(x+b)}{x^2+ax}[/tex]
It can be rewritten as
[tex]f(x)=\dfrac{(x+a)(x+b)}{x^2+ax} = \dfrac{(x+a)(x+b)}{x(x+a)}[/tex]
We can cancel out the common factor (x+a) on the condition that [tex]x \neq -a[/tex]
The function becomes not defined at
[tex]x = -a\\x= 0[/tex]
Thus, it is discontinued at those points.
Taking left and right sided limits, on x = -a and x = 0 of function f(x), we get:
[tex]lim_{x \rightarrow -a^-}(\dfrac{(x+a)(x+b)}{x^2+ax}) = lim_{x \rightarrow -a^-}(\dfrac{(x+b)}{x}) = \dfrac{a-b}{a}\\\\\\lim_{x \rightarrow -a^+}(\dfrac{(x+a)(x+b)}{x^2+ax}) = lim_{x \rightarrow -a^+}(\dfrac{(x+b)}{x}) = \dfrac{a+b}{a}\\\\\rm Thus,\\\\lim_{x\rightarrow -a^-}(f(x)) \neq lim_{x\rightarrow -a^+}(f(x))[/tex]
Due to this, the discontinuity at x = -a is not removable(intuitively, think of it as the function itself is not agreeing on a single point able to be replaced for x = -a).
[tex]lim_{x \rightarrow 0^-}(\dfrac{(x+a)(x+b)}{x^2+ax}) = lim_{x \rightarrow 0^-}(\dfrac{(x+b)}{x}) =\infty \\\\\\lim_{x \rightarrow 0^+}(\dfrac{(x+a)(x+b)}{x^2+ax}) = lim_{x \rightarrow 0^+}(\dfrac{(x+b)}{x}) = \infty \\\\\rm[/tex]
We cannot compare infinity as equal as its not a value but a sign that there is arbitrary large or small value coming as result of the operation.
Thus, the discontinuity here is also not removable(since limits are not finite)
But, here as the function is continuous in the neighbourhood of x = 0, and values are going infinity, thus, there can lie a vertical asymptote at x= 0.
Thus, we conclude that:
- There is a vertical asymptote at x = 0 for the given function.
- There are no removable discontinuity for the given function.
Learn more about vertical asymptotes here:
https://brainly.com/question/2513623
