Answer:
"a removable discontinuity that can be removed by extending the function to include f(1) = 13"
Step-by-step explanation:
We have the function:
[tex]f(x) = \frac{x^2 + 11*x - 12}{x - 1}[/tex]
Here we have a problem when x = 1, because that makes the denominator to be equal to zero.
For now, let's see what happens to the numerator when x = 1
1^2 + 11*1 - 12 = 12 - 12 = 0
So x = 1 is a root of the numerator.
Let's find the other root of the numerator, here we can use Bhaskara's formula:
[tex]x = \frac{-11 \pm \sqrt{11^2 - 4*1*(-12)} }{2*1} = \frac{-11 \pm 13}{2}[/tex]
Then the two roots are:
x = (-11 + 13)/2 = 1
x = (-11 - 13)/2 = -12
And remember that a quadratic equation:
y = a*x^2 + b*x + c
With roots p and k, can be written as:
y = a*(x - p)*(x - k)
Then we can rewrite our numerator as:
1*(x - 1)*(x - (-12)) = (x - 1)*(x + 12)
Replacing that in the equation for f(x), we get:
[tex]f(x) = \frac{x^2 + 11*x - 12}{x - 1} = \frac{(x - 1)*(x + 12)}{(x - 1)} = \frac{x - 1}{x -1} *(x + 12)[/tex]
f(x) = x + 12
And, when evaluated in x = 1, we get:
f(1) = 1 + 12 = 13
Then the correct option is:
"a removable discontinuity that can be removed by extending the function to include f(1) = 13"
