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inessss [21]
3 years ago
7

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

Mathematics
2 answers:
qwelly [4]3 years ago
6 0

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)

tatiyna3 years ago
3 0

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

<u>Step-by-step explanation:</u>

  \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

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