Piecewise Function

[nycmath]

See attachment.

[Math Tutor]

B. Domain. Take the union of the three pieces: [−4, 0) ∪ {0} ∪ (0, ∞) = [−4, ∞).

A. Range. Piece by piece:
• y = x on [−4, 0) gives [−4, 0)
• x = 0 gives the single value 1
• y = 3x for x > 0 gives (0, ∞)
Union: [−4, 0) ∪ (0, ∞) — that is, every y ≥ −4 except y = 0. (The value 1 is already in (0, ∞).)

C. Intercepts. y-intercept: f(0) = 1, so (0, 1). x-intercepts: none — on [−4, 0) we would need x = 0, which is not in that interval; for x > 0 we would need 3x = 0, again impossible; and f(0) = 1 ≠ 0.

D. Graph. The segment y = x from the closed point (−4, −4) to the open point (0, 0); the isolated point (0, 1); the ray y = 3x for x > 0, starting at the open point (0, 0) and rising steeply.

E. Continuity. On [−4, 0) and on (0, ∞) f is a straight line, hence continuous; at x = −4 it is continuous from the right (the endpoint of the domain). The only issue is x = 0:
lim(x→0−) f(x) = 0 and lim(x→0+) f(x) = 0, so lim(x→0) f(x) = 0, but f(0) = 1.
Since the limit exists but does not equal f(0), f is not continuous at x = 0 — a removable ("hole") discontinuity. So the answer to E is no: f is continuous everywhere on its domain except at x = 0.

[nycmath]

B. Domain. Take the union of the three pieces: [−4, 0) ∪ {0} ∪ (0, ∞) = [−4, ∞).

A. Range. Piece by piece:
• y = x on [−4, 0) gives [−4, 0)
• x = 0 gives the single value 1
• y = 3x for x > 0 gives (0, ∞)
Union: [−4, 0) ∪ (0, ∞) — that is, every y ≥ −4 except y = 0. (The value 1 is already in (0, ∞).)

C. Intercepts. y-intercept: f(0) = 1, so (0, 1). x-intercepts: none — on [−4, 0) we would need x = 0, which is not in that interval; for x > 0 we would need 3x = 0, again impossible; and f(0) = 1 ≠ 0.

D. Graph. The segment y = x from the closed point (−4, −4) to the open point (0, 0); the isolated point (0, 1); the ray y = 3x for x > 0, starting at the open point (0, 0) and rising steeply.

E. Continuity. On [−4, 0) and on (0, ∞) f is a straight line, hence continuous; at x = −4 it is continuous from the right (the endpoint of the domain). The only issue is x = 0:
lim(x→0−) f(x) = 0 and lim(x→0+) f(x) = 0, so lim(x→0) f(x) = 0, but f(0) = 1.
Since the limit exists but does not equal f(0), f is not continuous at x = 0 — a removable ("hole") discontinuity. So the answer to E is no: f is continuous everywhere on its domain except at x = 0.

Nicely-done. An easy to understand reply.