Answer:
It is continuous since [tex]\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)[/tex]
Step-by-step explanation:
We are given that the function is defined as follows [tex]f(x) = 9-x, x\leq 0[/tex] and [tex] f(x) = 9+12x, x>0[/tex] and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity) x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all [tex]\mathbb{R}[/tex]. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that
[tex]\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)[/tex] (this is the definition of continuity at x=0)
Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that
[tex]\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9[/tex]. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.
Note that when x>0, we have that f(x) = 9+12x. In this case, we have that
[tex]\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9[/tex]. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.
Thus, [tex]\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9[/tex], so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].
