L42 Solutions — Inverse and Composite Functions

1. Composite functions — basic. Fluency

   • (a) f(g(x)) = f(x−3) = 2(x−3)+1 = 2x − 5.
   • (b) g(f(x)) = g(2x+1) = (2x+1)−3 = 2x − 2.
   • (c) f(f(x)) = f(2x+1) = 2(2x+1)+1 = 4x+2+1 = 4x + 3.
   • (d) g(4)=1, so f(g(4))=f(1)=3. f(4)=9, so g(f(4))=g(9)=6. f(g(4))=3, g(f(4))=6 — they are different, confirming f∘g ≠ g∘f.

2. Finding inverses — linear. Fluency

   • (a) y=x+7 ⇒ swap ⇒ x=y+7 ⇒ y=x−7. f⁻¹(x) = x − 7.
   • (b) x=3y−5 ⇒ 3y=x+5 ⇒ y=(x+5)/3. g⁻¹(x) = (x+5)/3.
   • (c) x=(y+2)/4 ⇒ 4x=y+2 ⇒ y=4x−2. h⁻¹(x) = 4x − 2.
   • (d) x=(1−2y)/3 ⇒ 3x=1−2y ⇒ 2y=1−3x ⇒ y=(1−3x)/2. k⁻¹(x) = (1−3x)/2.

3. Inverse verification. Fluency

   • (a) f(g(x)) = 5⋅(x+3)/5 − 3 = (x+3)−3 = x ✓. g(f(x)) = (5x−3+3)/5 = 5x/5 = x ✓. They are inverses.
   • (b) The inverse reverses the mapping. Since f(7) = 4, the inverse maps 4 back to 7. f⁻¹(4) = 7.
   • (c) f⁻¹(2) = −3 means the inverse maps 2 to −3, so the original maps −3 to 2. f(−3) = 2.
   • (d) Find f by inverting f⁻¹: let y=2x−6, swap: x=2y−6, solve: y=(x+6)/2. f(x) = (x+6)/2.

4. Non-linear composites. Fluency

   • (a) p(q(x)) = p(x+4) = (x+4)² = x² + 8x + 16.
   • (b) q(p(x)) = q(x²) = x² + 4. (Not the same as (a).)
   • (c) q(−2) = −2+4 = 2. p(2) = 4. p(q(−2)) = 4.
   • (d) (x+4)² = 9 ⇒ x+4 = ±3 ⇒ x = −1 or x = −7.

5. Inverse function from a graph. Understanding

   • (a) Domain of f: x ≥ 1 (need x−1 ≥ 0). Range: √(x−1) ≥ 0, so f(x) ≥ 2. Range: y ≥ 2.
   • (b) y = √(x−1)+2 ⇒ swap ⇒ x = √(y−1)+2 ⇒ x−2 = √(y−1) ⇒ (x−2)² = y−1 ⇒ f⁻¹(x) = (x−2)²+1. The red parabola on the graph confirms this — it starts at (2,1) as expected.
   • (c) Domain of f⁻¹: x ≥ 2 (was range of f). Range of f⁻¹: y ≥ 1 (was domain of f).
   • (d) f(x) = f⁻¹(x) at points on the line y = x (the mirror line). Solving √(x−1)+2 = x: let u=√(x−1), then u+2=u²+1+1=u²+1... substitute x=u²+1: u+2=u²+1 ⇒ u²−u−1=0 ⇒ u=(1+√5)/2. Then x = u²+1 = (5+√5)/2 ≈ 3.62. These points lie on y = x.

6. Composite functions — mixed. Understanding

   • (a) f(g(x)) = √(x−4). Domain: need x−4 ≥ 0, so x ≥ 4. Range: y ≥ 0.
   • (b) g(f(x)) = √x − 4. Domain: need x ≥ 0, so x ≥ 0. Range: y ≥ −4.
   • (c) No — they have different rules (√(x−4) vs √x−4) and different domains ([4,∞) vs [0,∞)). Composition is not commutative in general.
   • (d) √(x−4) = 3 ⇒ x−4 = 9 ⇒ x = 13.

7. Inverse functions — non-linear. Understanding

   • (a) y=x³+1 ⇒ x=y³+1 ⇒ y³=x−1 ⇒ f⁻¹(x) = &cbrk;x−1 (cube root). Both f and f⁻¹ have domain and range = all reals.
   • (b) f is not one-to-one: f(2) = f(−2) = 7. Swapping x and y gives x = y²+3 ⇒ y = ±√(x−3), which is not a function. Restrict domain to x ≥ 0 (or x ≤ 0) to make f one-to-one and allow a function inverse.
   • (c) y=2/x ⇒ x=2/y ⇒ y=2/x. f⁻¹(x) = 2/x = f(x). The function is self-inverse — applying it twice returns to the original value.
   • (d) Domain of f⁻¹ = range of f = [1, 11]. Range of f⁻¹ = domain of f = [0, 5].

8. Composites with three functions. Understanding

   • (a) f(x)=x+1 ⇒ g(f(x))=2(x+1)=2x+2 ⇒ h(g(f(x)))=(2x+2)² = 4(x+1)² = 4x²+8x+4.
   • (b) h(x)=x² ⇒ g(h(x))=2x² ⇒ f(g(h(x)))=2x²+1. f(g(h(x))) = 2x² + 1.
   • (c) f(3)=4, g(4)=8, h(8)=64. h(g(f(3))) = 64.
   • (d) h(g(f(x)))=4(x+1)². g(f(h(x))): h(x)=x², f(x²)=x²+1, g(x²+1)=2x²+2. Set equal: 4(x+1)²=2x²+2 ⇒ 4x²+8x+4=2x²+2 ⇒ 2x²+8x+2=0 ⇒ x²+4x+1=0 ⇒ x = −2 ± √3.

9. Applied composition and inverse. Problem Solving

   • (a) g(f(x)) = g(x−5) = 3(x−5) = 3x − 15. This gives the number of units produced from x kg of raw material (after waste is removed and the remainder tripled).
   • (b) g(f(50)) = 3(50) − 15 = 150 − 15 = 135 units.
   • (c) Let h(x)=3x−15. Inverse: x=3y−15 ⇒ y=(x+15)/3. h⁻¹(x) = (x+15)/3. This represents the raw material needed to produce x units.
   • (d) h⁻¹(120) = (120+15)/3 = 135/3 = 45 kg of raw material.

10. Decomposing functions. Problem Solving

    • (a) h(x) = (3x+2)⁵. Let g(x) = 3x+2 (inner) and f(x) = x⁵ (outer). Then f(g(x)) = (3x+2)⁵ ✓.
    • (b) h(x) = √(x²+1). Let g(x) = x²+1 (inner) and f(x) = √x (outer). Then f(g(x)) = √(x²+1) ✓.
    • (c) Example: f(x) = 1/x. f(f(x)) = f(1/x) = 1/(1/x) = x ✓. Since f(f(x)) = x, applying f twice returns to the start — f is its own inverse: f = f⁻¹.
    • (d) f(g(x)) = 2g(x)+1 = 6x−3 ⇒ 2g(x) = 6x−4 ⇒ g(x) = 3x − 2.