Question

Identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or neither. 5x/ x3 + 5x2 + 6x

Answer 1

Answer:

Function is discontinuous at

x=0   (Hole)

x=-3  (Vertical asymptote)

x=-2  (Vertical asymptote)

Step-by-step explanation:

Given: $f(x)=\dfrac{5x}{x^3+5x^2+6x}$

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.

$f(x)=\dfrac{5x}{x^3+5x^2+6x}$

$f(x)=\dfrac{5x}{x(x^2+5x+6)}$

$f(x)=\dfrac{5x}{x(x+3)(x+2)}$

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)

Answer 2

Answer: hole at x = 0, asymptotes at x = -2 and x = -3

Step-by-step explanation:

  $\frac{5x}{x^{3} + 5x^{2} + 6x}$

= $\frac{5x}{x(x^{2}+5x + 6)}$

= $\frac{5x}{x(x+2)(x+3)}$

x ≠ 0 ---> the x cancels out so this is a hole  

x + 2 ≠ 0   ---> x ≠ -2   ---> asymptote at x = -2

x + 3 ≠ 0   ---> x ≠ -3   ---> asymptote at x = -3