B
Bob
Guest
Find all points of discontinuity.
f(x) = 16/x^2 if x>= 2 ; 3x-2 if x<2
I'm not sure if I'm correct... but I believe that there are no points of discontinuity for this function.
Here is how I figure,
For 16/x^2, the rational function has a denominator x^2 = (x+0)(x+0) which is zero when x=1. Thus, f would be discontinuous at 0; however, it is defined only for values of x >= 2, therefore, there are no discontinuities.
For 3x-2, this case defining the function is given by a polynomial, and polynomials are always continuous.
Am I overlooking anything... or am I correct?
Feedback would be greatly appreciated. Thank you.
f(x) = 16/x^2 if x>= 2 ; 3x-2 if x<2
I'm not sure if I'm correct... but I believe that there are no points of discontinuity for this function.
Here is how I figure,
For 16/x^2, the rational function has a denominator x^2 = (x+0)(x+0) which is zero when x=1. Thus, f would be discontinuous at 0; however, it is defined only for values of x >= 2, therefore, there are no discontinuities.
For 3x-2, this case defining the function is given by a polynomial, and polynomials are always continuous.
Am I overlooking anything... or am I correct?
Feedback would be greatly appreciated. Thank you.