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3 years ago

PLZZ HELPP ASAP MATH QUESTION SIMPLIFY COMPLEX FRACTION

Mathematics

1 answer:

Ksivusya [100]3 years ago

6 0

Answer:

$\large\boxed{\dfrac{x^3+6x}{80x+64}}$

Step-by-step explanation:

$\dfrac{\frac{1}{8}+\frac{x+4}{16}}{\frac{x+4}{x^2}+\frac{4}{x}}=\left(\dfrac{1\cdot2}{8\cdot2}+\dfrac{x+4}{16}\right):\left(\dfrac{x+4}{x^2}+\dfrac{4\cdot x}{x\cdot x}\right)\\\\=\left(\dfrac{2}{16}+\dfrac{x+4}{16}\right):\left(\dfrac{x+4}{x^2}+\dfrac{4x}{x^2}\right)=\dfrac{2+x+4}{16}:\dfrac{x+4+4x}{x^2}\\\\=\dfrac{x+6}{16}:\dfrac{5x+4}{x^2}=\dfrac{x+6}{16}\cdot\dfrac{x^2}{5x+4}=\dfrac{(x+6)(x^2)}{(16)(5x+4)}\\\\\text{use distributive property}\ a(b+c)=ab+ac$

$=\dfrac{(x)(x^2)+(6)(x)}{(16)(5x)+(16)(4)}=\dfrac{x^3+6x}{80x+64}$

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a) Since both limits are distinct and do not exist, we conclude that x = - 1 is not part of the domain of the rational function.

b) The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

<h3>How to determine whether a limit exists or not</h3>

According to theory of limits, a function f(x) exists for x = a if and only if $\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$ . This criterion is commonly used to prove continuity of functions.

Rational functions are not continuous for all value of x, as there are x-values that make denominator equal to 0. Based on the figure given below, we have the following lateral limits:

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The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

To learn more on lateral limits: brainly.com/question/21783151

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