Find value(s) of parameter c (if exists) that make f continuous everywhere.

Question

thesmartone · Answer

$\Large f(x) = \frac{2}{x^2+4x+c}$

thesmartone · Answer

There was a question before it with the same instructions
$\Large f(x) = \frac{2}{x+c}$

I feel like the answers to both of them are there is no c value that will make f continuous everwhere.

unklerhaukus · Answer

Is there some restriction on the domain of f ?

thesmartone · Answer

No

unklerhaukus · Answer

what are the solution to
 x^2+4x+c = 0 ?

when are these solutions not real numbers?

thesmartone · Answer

x^2 + 4x + 4 = 0
(x + 2)^2 = 0
x = -2

c = (-2)^2 = 4

so whenever c > 4 the solutions to it will not be real numbers

unklerhaukus · Answer

when c>4 the denominator cannot be zero

thesmartone · Answer

I made c = 4 (because perfect square trinomials)  and then calculated x and then found the value of c again and I'm going crazy x'D

But when c > 4 the solutions will not be real numbers

thesmartone · Answer

Ohhhh o:

That makes a lot of sense! Thanks Unkle! :D