Question

Explain why f(x) = √x2+4x+3/x2-x-2 is not continuous at x = -1.

Answer 1

Answer: Option C

Step-by-step explanation:

Make the denominator equal to zero and solve for x:

$x^{2}-x-2=0\\(x-2)(x+1)=0\\x=2\\x=-1$

Then, as you can see, x=-1 makes the denominator equal to zero and the division by zero does not exist Therefore you can conclude that the function shown in the problem is not defined at x=-1

The answer is the option C.

Answer 2

Answer:

Option C. f(x) is not defined at x = -1

Step-by-step explanation:

Since given function is $f(x) = \frac{x^{2}+4x+3 }{x^{2} -x-2}$

We have to check the continuity of the given function at x = -1.

We rewrite the function in the form an equation

$y = \frac{x^{2}+4x+3 }{x^{2} -x-2}$

Now we factorize the fraction of the expression

$y=\frac{x^{2}+4x+3}{x^{2}-x-2}$

$=\frac{x^{2}+3x+x+3}{x^{2}-2x+x-2}$

$=\frac{(x+1)(x+3)}{(x-2)(x+1)}$

Now we can explain that for the value of x from the denominator

(x -2) = 0 Or x = 2

and for (x +1) =0

Or x = -1

The function is not continuous.

Therefore Option C is the correct answer.