Answer
$a = \frac{1}{2}- \frac{\sqrt{5}}{2}$
$a = \frac{1}{2}+ \frac{\sqrt{5}}{2}$
Work Step by Step
For $f$ to be continuous on the real numbers, it is required that $\lim\limits_{x \to a}x^2 = a+1$
Then:
$a^2=a+1$
$a^2-a-1 = 0$
We can use the quadratic formula to find the values of $a$:
$a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$a = \frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{(2)(1)}$
$a = \frac{1 \pm \sqrt{5}}{2}$
$a = \frac{1}{2}- \frac{\sqrt{5}}{2}~~~$ or $~~~a = \frac{1}{2}+ \frac{\sqrt{5}}{2}$
