Piecewise and Continuity?

Question

Please help:
Find all values of a such that f is continuous on all real numbers:

f(x)={x+1 when x is less than or equal to a
{x^2 when x>a

Answer 1

Since both "pieces" are polynomials, their domains are both all real numbers. So all we have to do is set the two equations equal to each other (basically you are connecting two continuous functions at the point a - graph it out to see what I mean).

x + 1 = x^2
x^2 - x - 1 = 0

Quadratic formula:
x = [1 +/- sqrt(1 - 4(-1))] / 2
x = [1 +/- sqrt(5)] / 2

So if a = [1 +/- sqrt(5)] / 2, the function is continuous.

Answer 2

Well, we know f(x) is continuous for all x ≠ a, so we have to find all values of a that make f(x) continuous at a.

To be continuous, the left and right hand limits must be equal. The left hand limit is lim(x->a-) f(x) = lim(x->a-) x+1 = a+1.
The right hand limit is lim(x->a+) f(x) = lim(x->a+) x^2 = a^2.
Also, the value at a is the left hand limit.
So f(x) will be continuous at a if and only if a^2 = a+1
<=> a^2 - a - 1 = 0
<=> a = (1 ± √5) / 2.

Answer 3

Let's say that x^2 and x+1 both approach a y-coordinate of a point.

x^2 = x + 1

x^2 - x - 1 = 0

1 +/- sqrt (1 + 4) / 2

1 +/- sqrt 5 / 2

Those are the two values of a that work.