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Elis [28]
2 years ago
14

1 and -1 is nonlinear or linear

Mathematics
1 answer:
OLga [1]2 years ago
4 0

Answer:

liner

Step-by-step explanation:

they look like a line on a graph

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F(x)
mina [271]

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

<em>Rational</em> 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 <em>lateral</em> limits:

\lim_{x \to -1^{-}} \frac{x}{x^{2}+x} = - \infty

\lim_{x \to -1^{+}} \frac{x}{x^{2}+x} = + \infty

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

In addition, we can simplify the function by <em>algebra</em> properties:

\frac{x}{x^{2}+ x} = \frac{x}{x\cdot (x + 1)} = \frac{1}{x + 1}

g(x) = \frac{1}{x + 1}

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

#SPJ1

7 0
2 years ago
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