Question

given the function f(x)= (square root x^2 -4)/ x-4, what values of x will the function be discontinuous. (explain your reasoning). thanks.

Answer 1

f(x) = sqrt(x^2 - 4) / (x-4)

A rationals denom cannot be 0, therefore, x cannot = 4

Therefore, the function is discontinous at x = 4.

Answer 2

so there is only one that is discontinuous?

Answer 3

Ahh sorry, the points at which it is discontinous are X < 2 and X = 4

Sqrt(x^2-4) cannot be less than 2, we cant take the sqrt of a negative.

Answer 4

how do you know that square root part cannot be less than 2?

Answer 5

ask yourself, can you take the square root of a negative number?

Say x is 1.

then sqrt( 1 - 4) = sqrt(-3) which is not defined (only in complex plane)

Answer 6

i typed it on my graphing calculator and it said that x=-1,0,1 are errors.

Answer 7

Exactly, it's discontinous at those points.

Answer 8

X has to be greater than 2, and cannot equal 4. This is the domain of the function.

I.e, it is discontinous at X < 2 and x=4

Answer 9

how did you can the 2 coming from? because when i sub in the 2 in the function it equals to 0

Answer 10

X LESS than 2 is different than X lesss than or EQUAL to 2.

X < 2 means 0,1 but not 2.
X <=2 means 0,1 AND 2 as well.

So going back to the question,

ANY value less than 2, is not defined because it makes our numerator (our square root), negative.

Answer 11

oh... i see. thanks lots for explaining that to me. :)

Answer 12

can you help me with another question please? i just starting calculus and I'm strugging doing it.