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3 years ago

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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3 years ago

Need help ASAP

joja [24]

Part (1) : The solution is $729$

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Explanation:

Part (1): The expression is $3^{2} \cdot3^{4}$

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$ , we get,

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Adding the exponent, we get,

$3^{2} \cdot3^{4}=3^6=729$

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Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

$\frac{1}{2^4x^8}$

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$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$ , we have,

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Adding the powers, we get,

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Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\frac{2 x^{2}}{3 y z^{7}}$

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