Question

Determine all numbers at which the function is continuous.

Answer 1

Answer: Option a.

Step-by-step explanation:

Make the denominator equal to zero and solve for x, as following:

$x^{2}+8x-9=0$

To solve the quadratic equation you can factor ir. You must  find two numbers whose sum is 8 and whose product is 9. These would be -1 and 9.

Then you have:

$(x-1)(x+9)=0\\x=1\\x=-9$

Therefore, based on this, you can conclude that the function is continuous at everty real number except x=1 and x=-9.

Answer 2

Answer:

Option A. is the correct option.

Step-by-step explanation:

In this question the given function is

$f(x)=\frac{x^{2}+5x-36}{x^{2}+8x-9}$

We have to find the continuity of the given function

If we rewrite the function in the factorial form

$f(x)=\frac{x^{2}+9x-4x-36}{x^{2}+9x-x-9}$

$=\frac{x(x+9)-9(x+9)}{x(x+9)-1(x+9)}$

$f(x)=\frac{(x+9)(x-9)}{(x+9)(x-1)}$

Now we take the denominator of the function

(x + 9) = 0

x = -9

and (x -1) = 0

x = 1

So for x = -9 and x = 0 the function becomes undefined.

Therefore function is continuous for every real number except x = -9 and x = 1.

Option a is the answer.