Brent K. answered • 06/07/21
Applied Math PhD
The formatting is a little off, so I'll try and be clear about what functions I'm considering, so you can adapt if needed.
The key principal is that elementary functions - i.e. essentially all the functions you can write a 'single' formula for - are continuous on their domains. Thus, if you can write a single formula for the function (so, piecewise defined functions are a possible exception where the rule changes), and you can input the given value, it is continuous at that value.
a) f(x) = (x^2-5)/(x)
This has domain all real numbers, except x=0 (you cannot divide by 0). Hence, it is continuous everywhere except x=0
b) f(x) = (3x-1)/(2x^2+6x)
You cannot divide by 0, and so determine where the denominator is equal to 0: factoring, 2x^2+6x=2x(x+3), which has roots at x=0 and x=-3. Therefore, f is continuous everywhere except x=0 and x=-3.
c)f(x) = sqrt(x^2-4)
You cannot take the square root of a negative number, and so we must have x^2-4 ≥0. This has solution set (-∞,-2] U [2,∞), which also gives where f is continuous (you can solve this inequality by factoring, or by graphing)
d) f(x) = (sqrt(x+3))/(x-2)
You cannot take the square root of a negative number, and you cannot divide by 0. For the first point, we must have x+3 ≥0, so x≥-3, which has solution set [-3, ∞). For the second point, we cannot have x=2.
Thus, the domain of f is [-3,2)U(2,∞).
e) f(x) = tan(x) = (sin(x))/(cos(x))
You cannot divide by 0, so we cannot have cos(x) = 0. This happens at pi/2, 3pi/2, .... and -pi/2, -3pi/2, ....
Generally, this can be expressed as pi/2 + kpi, where k is an arbitrary integer. f is continuous at all points except those.
