Question

Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The function is discontinuous because all piecewise functions are discontinuous at their domain boundaries.The function is discontinuous because f(−4) ≠ f(5). The function is continuous because lim x→0− f(x) = lim x→0+ f(x) = f(0) = 9.The function is continuous because the domain boundary is not inside the interval [−4, 5].The function is continuous because lim x→−4+ f(x) and lim x→5− f(x) both exist.

Answer 1

Answer:

Continuous

Step-by-step explanation:

Answer 2

Answer:

It is continuous since $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$

Step-by-step explanation:

We are given that the function is defined as follows $f(x) = 9-x, x\leq 0$ and $f(x) = 9+12x, x>0$ 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 $\mathbb{R}$. 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

$\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)$ (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

$\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9$. 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

$\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9$. 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, $\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9$, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].