It looks like the given function is
[tex]f(x) = \begin{cases}-3x + 7 & \text{if }x < 0 \\ x^2+7 & \text{if }x \ge0\end{cases}[/tex]
The two pieces of f(x) are continuous since they are polynomials, so we only need to worry about the point at which they meet, x = 0. f(x) is continuous there if
[tex]\displaystyle \lim_{x\to0^-} f(x) = \lim_{x\to0^+} f(x) = f(0) = 7[/tex]
To the left of x = 0, we have x < 0, so f(x) = -3x + 7 :
[tex]\displaystyle \lim_{x\to0^-} f(x) = \lim_{x\to0} (-3x + 7) = 7[/tex]
To the right of x = 0, we have x > 0 and f(x) = x² + 7 :
[tex]\displaystyle \lim_{x\to0^+} f(x) = \lim_{x\to0} (x^2 + 7) = 7[/tex]
So f(x) is continuous at x = 0, and hence continuous for all real numbers.
