Answer:
x = −3 asymptote
x = −2 asymptote
x = 0 hole
x = 2 neither
x = 3 neither
x = 5 neither
Step-by-step explanation: edg 2020
A discontinuous function is simply the opposite of a continuous function.
The values of x are:
The function is given as:
[tex]\mathbf{f(x) = \frac{5x}{x^3 + 5x^2 + 6x}}[/tex]
Factor out x
[tex]\mathbf{f(x) = \frac{5x}{x(x^2 + 5x + 6)}}[/tex]
Cancel out the common factor
[tex]\mathbf{f(x) = \frac{5}{x^2 + 5x + 6}}[/tex]
So, we have:
[tex]\mathbf{x = 0}[/tex] --- hole
Expand
[tex]\mathbf{f(x) = \frac{5}{x^2 + 2x + 3x + 6}}[/tex]
Factorize
[tex]\mathbf{f(x) = \frac{5}{x(x + 2) + 3(x + 2)}}[/tex]
Factor out x + 2
[tex]\mathbf{f(x) = \frac{5}{(x + 3)(x + 2)}}[/tex]
Equate the denominator to 0
[tex]\mathbf{(x + 3)(x + 2) = 0}[/tex]
Split
[tex]\mathbf{x + 3 = 0 \ or \ x + 2 = 0}[/tex]
Solve for x
[tex]\mathbf{x = -3 \ or \ x = -2}[/tex] --- asymptote.
So, the other x values are none.
Read more about discontinuity of functions at:
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