Question

X^2-16/(x+4)(x-5) x=-4 x=1 continuous at x=-4?

Answer 1

Answer:

Yes, its continuous

Step-by-step explanation:

We use the formula:

x^2-y^2=(x-y)(x+y),

And we know that 16=4^2, so we have:

$\frac{x^2-16}{(x+4)(x-5)}=\frac{(x-4)(x+4)}{(x+4)(x-5)}=\frac{x-4}{x-5}$

So for x=-4 we have -8/-9,i.e, it is 8/9, so it is continuous.

I dont know what is x=1, because for x=1 the function has value 3/4.

But function is not continuous in x=5 becaus for that x we will get 1/0, and that is not definite.

:)