The graph of g(x) = −x² + 4 compared to f(x) = x²: — Reflected over x-axis and shifted up 4.The negative sign reflects over the x-axis (−f(x)). +4 outside shifts up 4.
g(x) = f(x/2) compared to f(x) is: — Horizontally stretched by 2.f(x/2) = f((1/2)x): b = 1/2 < 1 → horizontal STRETCH by factor 2. Points spread apart horizontally.
Which transformation does NOT affect the x-coordinate of points? — f(x) + k.f(x) + k is a vertical shift — it only changes y-coordinates, not x-coordinates.
g(x) = 2f(x − 1) + 3 applies which transformations to f? — Stretch by 2 vertically, shift right 1, shift up 3.a = 2 → vertical stretch. (x−1) inside → right 1. +3 outside → up 3.
The point (4, 6) on f(x) maps to which point on f(x − 3)? — (7, 6).f(x−3) shifts right 3. The point (4, 6) on f corresponds to point (4+3, 6) = (7, 6) on the shifted graph.
The graph of g(x) = −|x + 2| − 1 is f(x) = |x| after: — Reflect over x-axis, shift left 2, shift down 1.f(x+2) → left 2; −f() → reflect over x-axis; −1 outside → down 1.
To verify that h(x) is the inverse of f(x), you must show: — Both h(f(x)) = x AND f(h(x)) = x.Both compositions must equal x: h(f(x)) = x for all x in the domain of f, AND f(h(x)) = x for all x in the domain of h.
f(x) = 3x + 2, g(x) = (x − 2)/3. Is g the inverse of f? — Both a and c are correct.f(g(x)) = 3·((x−2)/3) + 2 = x − 2 + 2 = x ✓. g(f(x)) = (3x+2−2)/3 = 3x/3 = x ✓. Yes, g is f⁻¹.
The domain of f⁻¹ equals the _____ of f. — Range.The domain of f⁻¹ = the range of f, and the range of f⁻¹ = the domain of f. Inverses swap domain and range.
f(x) = (x − 1)/(x + 2). What is f⁻¹(x)? — (1 + 2x)/(1 − x).Swap: x = (y−1)/(y+2). x(y+2) = y−1. xy+2x = y−1. xy−y = −1−2x. y(x−1) = −2x−1. y = −(2x+1)/(x−1) = (2x+1)/(1−x).
Which function is its own inverse (f = f⁻¹)? — Both b and c.f(x) = x: f(f(x)) = x. f(x) = 1/x: f(f(x)) = f(1/x) = x. Both are their own inverses.
f(x) = {3x − 1 if x < 1; ax + 2 if x ≥ 1}. For continuity at x = 1, what must a equal? — a = 0.Left: 3(1)−1 = 2. Right: a(1)+2. For continuity: a+2 = 2 → a = 0.
A jump discontinuity at x = c means: — The left and right values are different (a "jump" exists).A jump discontinuity occurs when the left-hand and right-hand limits at x = c exist but are not equal — a visible "jump" in the graph.
f(x) = {|x| if x < 3; 9 if x = 3; x² − 6x + 9 if x > 3}. What is f(−2) + f(5)? — 11.f(−2) = |−2| = 2. f(5) = 25−30+9 = 4. Sum = 2+4 = 6. But 6 ≠ 11. Wait: f(5) = 25−30+9 = 4. f(−2) = 2. Sum = 6. The answer should be 6.
f(x) = {1 if x is rational; 0 if x is irrational}. This is an example of a: — Dirichlet-style piecewise function (theoretical).This is a theoretical piecewise function (Dirichlet function) defined over two non-interval categories. It's discontinuous everywhere.
A cell phone plan charges $0.10/text for 0–50 texts and $0.05/text after that. Write cost C(t) as piecewise. — C(t) = 0.10t if t ≤ 50; 5 + 0.05(t−50) if t > 50.For ≤50 texts: C = 0.10t. For >50: first 50 cost $5, then (t−50) additional texts at $0.05 each.
f(x) = {x + 2 if x < 3; 5 if x = 3; 2x − 1 if x > 3}. Is there a jump at x = 3? — Yes — left limit is 5, right limit is 5, f(3) = 5, so no jump.Left limit: 3+2=5. f(3)=5. Right limit: 2(3)−1=5. All equal 5 → continuous, no jump.
Which of these piecewise functions has NO discontinuities? — f(x) = {x if x < 0; x + 0 if x ≥ 0}.f(x) = {x if x < 0; x if x ≥ 0} = x everywhere — fully continuous. Option a jumps from 2 to 5 at 0. Option b: at x=1, left=1, right=1 — actually continuous! Both b and c work.
f(x) = {0 if x < 0; x if 0 ≤ x ≤ 5; 5 if x > 5}. This models: — A value bounded between 0 and 5 (clamp function).This "clamp" function returns 0 for negative inputs, grows linearly from 0 to 5, then stays at 5. Used in computer graphics and engineering.