4 years ago

What’s the squareroot -1?

Mathematics

1 answer:

Murrr4er [49]4 years ago

6 0

Answer:

You can't calculate the square root of -1

Step-by-step explanation:

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Explain why f(x) = x^2-x-6/x^2-9 is not continuous at x = 3.

iren [92.7K]

Answer:

See Explanation

Step-by-step explanation:

Given

$f(x) = \frac{x^2 - x -6}{x^2 - 9}$

Required

Why is the function not continuous at x = 3

First substitute 3 for x at the denominator

$f(x) = \frac{x^2 - x -6}{x^2 - 9}$

Factorize the numerator and the denominator

$f(x) = \frac{x^2 - 3x+2x -6}{x^2 - 3^2}$

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

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

Divide the numerator and denominator by (x - 3)

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

Substitute 3 for x

$f(3) = \frac{3+2}{3+3}$

$f(3) = \frac{5}{6}$

Because $f(x) = \frac{x^2 - x -6}{x^2 - 9}$ is defined when x = 3;

Then the function is continuous

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3 years ago

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3 years ago