Answer:
Function is discontinuous at
x=0 (Hole)
x=-3 (Vertical asymptote)
x=-2 (Vertical asymptote)
Step-by-step explanation:
Given: [tex]f(x)=\dfrac{5x}{x^3+5x^2+6x}[/tex]
We need to identity the discontinuity of the function. As we know function is discontinuous where it is not defined.
So, The function is discontinuous at hole, asymptote and break point.
[tex]f(x)=\dfrac{5x}{x^3+5x^2+6x}[/tex]
[tex]f(x)=\dfrac{5x}{x(x^2+5x+6)}[/tex]
[tex]f(x)=\dfrac{5x}{x(x+3)(x+2)}[/tex]
For hole, we will cancel like factor from numerator and denominator.
At x=0 we get hole.
For vertical asymptote, we set denominator to 0
x+3=0 and x+2=0
Vertical asymptote:
x=-3 and x=-2
Function is discontinuous at
x=0 (Hole)
x=-3 (Vertical asymptote)
x=-2 (Vertical asymptote)
The vertical asymptote of a function are the zeroes of the denominator of a rational function
The function is discontinuous at:
The function is given as:
[tex]\mathbf{f(x) = \frac{5x}{x^3 + 5x^2 + 6x}}[/tex]
Factor out x, from the denominator
[tex]\mathbf{f(x) = \frac{5x}{x(x^2 + 5x + 6)}}[/tex]
Expand
[tex]\mathbf{f(x) = \frac{5x}{x(x^2 + 2x +3x + 6)}}[/tex]
Factorize
[tex]\mathbf{f(x) = \frac{5x}{x(x(x + 2) +3(x + 2)}}[/tex]
Factor out x + 2
[tex]\mathbf{f(x) = \frac{5x}{x(x + 2)(x +3)}}[/tex]
Cancel out like factors
[tex]\mathbf{f(x) = \frac{5}{(x + 2)(x +3)}}[/tex]
There is no coefficient of x in the numerator.
So, we have:
[tex]\mathbf{x = 0}[/tex] --- hole
Equate the denominator of [tex]\mathbf{f(x) = \frac{5}{(x + 2)(x +3)}}[/tex] to 0
[tex]\mathbf{x + 2 =0\ or\ x +3= 0}}[/tex]
Solve for x
[tex]\mathbf{x =-2\ or\ x = -3}}[/tex]
So, the vertical asymptote is:
[tex]\mathbf{x =-2\ or\ x = -3}}[/tex]
Read more about vertical asymptote and holes at:
https://brainly.com/question/9404499
