FUN-1 — Solutions Reference

P1 — Which Correctly Represents f(x+h)? (5 Variants)

The correct answer for each variant and explanation of common errors:

Variant 0 — f(x) = 5x − 2

Correct answer: \( 5x + 5h - 2 \)

Substitute (x+h) for every x: \( f(x+h) = 5(x+h) - 2 = 5x + 5h - 2 \)

Variant 1 — f(x) = −3x + 7

Correct answer: \( -3x - 3h + 7 \)

\( f(x+h) = -3(x+h) + 7 = -3x - 3h + 7 \)

Variant 2 — f(x) = x² + 6

Correct answer: \( x^2 + 2xh + h^2 + 6 \)

\( f(x+h) = (x+h)^2 + 6 = x^2 + 2xh + h^2 + 6 \)

Common error: \( x^2 + h^2 + 6 \) — Freshman's Dream, missing the \( 2xh \) cross term.

Variant 3 — f(x) = 3x² − 2x

Correct answer: \( 3x^2 + 6xh + 3h^2 - 2x - 2h \)

\[ f(x+h) = 3(x+h)^2 - 2(x+h) = 3(x^2+2xh+h^2) - 2x - 2h = 3x^2+6xh+3h^2-2x-2h \]

Variant 4 — f(x) = 4 − 2x²

Correct answer: \( 4 - 2x^2 - 4xh - 2h^2 \)

\[ f(x+h) = 4 - 2(x+h)^2 = 4 - 2(x^2+2xh+h^2) = 4 - 2x^2 - 4xh - 2h^2 \]

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P2 — Computing (f∘g)(x) Step by Step

\( f(x) = x^2 - 5 \), \( g(x) = 3x + 1 \)

Step 1 — Inner function: g(x) = 3x + 1 (g runs first in f∘g)

Step 2 — Substitution: Replace x in f with (3x + 1)

Step 3 — Unsimplified form: \( f(3x+1) = (3x+1)^2 - 5 \)

Step 4 — Simplified form:

\[ (3x+1)^2 - 5 = 9x^2 + 6x + 1 - 5 = 9x^2 + 6x - 4 \]

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P3 — Decomposition (5 Variants)

Variant 0 — h(x) = (x² + 3)⁵

Inner: \( g(x) = x^2 + 3 \)  |  Outer: \( f(u) = u^5 \)

Variant 1 — h(x) = √(x − 4)

Inner: \( g(x) = x - 4 \)  |  Outer: \( f(u) = \sqrt{u} \)

Variant 2 — h(x) = (5x + 2)³

Inner: \( g(x) = 5x + 2 \)  |  Outer: \( f(u) = u^3 \)

Variant 3 — h(x) = (x² − 1)⁴

Inner: \( g(x) = x^2 - 1 \)  |  Outer: \( f(u) = u^4 \)

Variant 4 — h(x) = √(2x² + 7)

Inner: \( g(x) = 2x^2 + 7 \)  |  Outer: \( f(u) = \sqrt{u} \)

IP1 — Compute f(x+h) (5 Variants)

Variant 0 — f(x) = 3x + 1

\[ f(x+h) = 3(x+h) + 1 = 3x + 3h + 1 \]

Difference: \( f(x+h) - f(x) = 3h \). In BTC-2, dividing by h gives slope = 3.

Variant 1 — f(x) = x² − 4x

\[ f(x+h) = (x+h)^2 - 4(x+h) = x^2 + 2xh + h^2 - 4x - 4h \]

Variant 2 — f(x) = 2x² + x − 3

\[ f(x+h) = 2(x+h)^2 + (x+h) - 3 = 2x^2 + 4xh + 2h^2 + x + h - 3 \]

Variant 3 — f(x) = 7 − 3x

\[ f(x+h) = 7 - 3(x+h) = 7 - 3x - 3h \]

Difference: \( f(x+h) - f(x) = -3h \). Slope = \( -3 \).

Variant 4 — f(x) = x³

\[ f(x+h) = (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \]

Using the binomial cube: \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).

Difference: \( f(x+h)-f(x) = 3x^2h + 3xh^2 + h^3 = h(3x^2 + 3xh + h^2) \). As \( h \to 0 \), this gives derivative \( 3x^2 \).

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IP2 — Generator: Evaluate (f∘g)(n)

The generator uses \( f(x) = x^2 + a \) and \( g(x) = bx + c \) with random parameters. The method is always the same:

1. Compute \( g(n) = bn + c \) — the intermediate value.
2. Compute \( f(g(n)) = (g(n))^2 + a \) — the final value.

Example with a = 5, b = 3, c = 2, n = 2:

\( g(2) = 3(2) + 2 = 8 \). Then \( f(8) = 64 + 5 = 69 \).

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IP3 — Given h(x), Identify f and g (5 Variants)

Variant 0 — h(x) = (x + 5)³

Inner: \( g(x) = x + 5 \)  |  Outer: \( f(u) = u^3 \)

Variant 1 — h(x) = √(x² + 1)

Inner: \( g(x) = x^2 + 1 \)  |  Outer: \( f(u) = \sqrt{u} \)

Variant 2 — h(x) = (2x − 3)⁴

Inner: \( g(x) = 2x - 3 \)  |  Outer: \( f(u) = u^4 \)

Variant 3 — h(x) = 1/(x² + x + 1)

Inner: \( g(x) = x^2 + x + 1 \)  |  Outer: \( f(u) = \dfrac{1}{u} \)

Domain note: \( x^2 + x + 1 \) has discriminant \( 1 - 4 = -3 < 0 \), so it is never zero — no real domain restrictions.

Variant 4 — h(x) = (x³ − 2x)²

Inner: \( g(x) = x^3 - 2x \)  |  Outer: \( f(u) = u^2 \)

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IP4 — Difference Quotient Preview

\( f(x) = x^2 + 1 \). Compute \( f(x+h) - f(x) \) and factor.

\[ f(x+h) = (x+h)^2 + 1 = x^2 + 2xh + h^2 + 1 \]

\[ f(x+h) - f(x) = (x^2 + 2xh + h^2 + 1) - (x^2 + 1) = 2xh + h^2 = h(2x + h) \]

MC1 — Feynman Test (Model Explanation)

\( f(x+h) \) means "substitute \( x+h \) as the entire input to f." \( f(x)+h \) means "evaluate f at x, then add h to the output." These are different operations in different order.

For \( f(x) = x^2 \): \( f(x+h) = (x+h)^2 = x^2 + 2xh + h^2 \), but \( f(x)+h = x^2 + h \). The difference is \( 2xh + h^2 - h \), which is not zero in general. The cross term \( 2xh \) is what carries the rate-of-change information.

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MC2 — Apply

\( f(x) = x^3 \), \( g(x) = 2x-5 \).

(f∘g)(2): \( g(2) = -1 \), then \( f(-1) = -1 \). Answer: \( -1 \).

(g∘f)(2): \( f(2) = 8 \), then \( g(8) = 11 \). Answer: \( 11 \).

Comparison: −1 ≠ 11. In general, composition is not commutative — swapping the order produces a different result.

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MC3 — Carlos's Error

Error identified: Carlos applied the Freshman's Dream: \( (x+h)^2 \neq x^2 + h^2 \). He dropped the cross term.

Correct computation:

\[ f(x+h) = (x+h)^2 - 3(x+h) = x^2 + 2xh + h^2 - 3x - 3h \]

Path A — The Analyst

\( f(x)=x+2 \), \( g(x)=3x-1 \), \( h(x)=x^2 \).

1. \( (f \circ g)(x) = f(3x-1) = 3x - 1 + 2 = 3x + 1 \)

2. \( (g \circ f)(x) = g(x+2) = 3(x+2)-1 = 3x+5 \)

3. \( h(2)=4 \), \( g(4)=11 \), \( f(11)=13 \). So \( (f \circ g \circ h)(2) = 13 \).

4. \( f(x+k)-f(x) = (x+k+2)-(x+2) = k \). The difference is just k — slope is 1.

5. Both are lines of slope 3: \( 3x+1 \) and \( 3x+5 \) are parallel. Setting them equal gives \( 1=5 \) — impossible. The compositions are never equal; they differ by a constant gap of 4.

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Path B — The Architect

\( p(x) = (2(x^2+1)+3)^3 \)

1. Triple decomposition:

• \( r(x) = x^2 + 1 \) (innermost)
• \( g(u) = 2u + 3 \) (middle)
• \( f(v) = v^3 \) (outer)

Verify: \( g(r(x)) = 2(x^2+1)+3 = 2x^2+5 \), then \( f(2x^2+5) = (2x^2+5)^3 = p(x) \quad ✓ \)

2. At x = 0: \( r(0)=1,\, g(1)=5,\, f(5)=125 \). Direct: \( (2(0+1)+3)^3 = 5^3 = 125 \quad ✓ \)

3. Two-function version: Merge r and g: \( a(x) = 2(x^2+1)+3 = 2x^2+5 \). Then \( p = f \circ a \) with \( f(v) = v^3 \).

4. Best for calculus: Three-function is better. The chain rule is applied one layer at a time: differentiate outer (3v²), then middle (2), then inner (2x) — three simple multiplications. The two-function version requires differentiating \( 2x^2+5 \) directly, which is less transparent.

C1 — When Composition Is Commutative

Example: \( f(x)=3x \) and \( g(x)=5x \). Both give \( 15x \).

Why: Scalar multiples through the origin satisfy \( f \circ g = fg \cdot x \) and multiplication of real numbers commutes.

General pattern: Any family of functions closed under composition and for which composition reduces to multiplication of a single parameter (like \( ax \) or \( x^n \)) will commute.

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C2 — Involutions

Involution 1: \( f(x)=-x \). Check: \( f(f(x))=f(-x)=x \quad ✓ \)

Involution 2: \( f(x)=a-x \) for any constant a. Check: \( f(f(x))=a-(a-x)=x \quad ✓ \)

Verification for 1/x: \( f(f(x))=f(1/x)=1/(1/x)=x \quad ✓ \) (domain: \( x \neq 0 \))

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C3 — Triple Decomposition via Completing the Square

\( h(x) = \sqrt{x^2+4x+3} \).

Complete the square: \( x^2+4x+3 = (x+2)^2 - 1 \).

Decomposition:

• \( r(x) = x+2 \)
• \( g(u) = u^2-1 \)
• \( f(v) = \sqrt{v} \)

Verify: \( g(r(x))=(x+2)^2-1=x^2+4x+3 \), then \( f(x^2+4x+3) = \sqrt{x^2+4x+3} = h(x) \quad ✓ \)