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Algebra 2 · Sec. 3.3 & 3.4

Operations on Functions 📐

Composition · Arithmetic · Inverse Functions
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Part 1 · Composition of Functions INSIDE-OUT
🔑 "INSIDE-OUT" f(g(x)) → Always evaluate the INNER function FIRST, then plug the result into the OUTER function.
Think: "Peel the onion from the inside."
f(g(x)) ≠ g(f(x)) — order matters! This is NOT commutative.
✏️ Example — How to evaluate h(f(x))
Given f(x) = x² − 3, h(x) = √(x + 2)
Step 1 → Find f(x) first: f(x) = x² − 3
Step 2 → Plug into h: h(f(x)) = h(x² − 3) = √((x² − 3) + 2) = √(x² − 1)
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Given: f(x) = x² − 3,   g(x) = 5x − 1,   h(x) = √(x + 2)
Evaluate f(3) — what is the value?
MY WORK
2
Given: f(x) = x² − 3,   g(x) = 5x − 1,   h(x) = √(x + 2)
Evaluate h(f(4)) — work inside-out!
MY WORK
3
Given: f(x) = x² − 3,   g(x) = 5x − 1,   h(x) = √(x + 2)
Evaluate g(h(14)) — what is the answer?
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⭐⭐
Given: f(x) = x² − 3,   g(x) = 5x − 1
Find the composed function g(f(x)) — simplify fully.
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⭐⭐
Given: f(x) = x² − 3,   g(x) = 5x − 1
Find f(g(x)) and simplify. Careful — order is reversed from #4!
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⭐⭐
Given: f(x) = x² − 3
Find f(f(x)) — a function composed with itself!
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⭐⭐
Given: g(x) = 5x − 1
Find g(g(x)) — compose g with itself and simplify.
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Part 2 · Arithmetic of Functions COMBINE
🔑 "LIKE TERMS ONLY" When adding or subtracting functions with same exponents, just combine coefficients.
e.g., 7x^(½) + (−x^(½)) = 6x^(½)just like 7a − a = 6a!
For multiplication: ADD the exponents → x^(a) · x^(b) = x^(a+b)
✏️ Example — Adding functions
f(x) = 7x^(1/2) − 2,   g(x) = −x^(1/2) + 4
f(x) + g(x) = (7x^(1/2) − x^(1/2)) + (−2 + 4) = 6x^(1/2) + 2
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f(x) = 7x^(1/2) − 2,   g(x) = −x^(1/2) + 4,   h(x) = −4x^(1/2) + 1
Find f(x) + g(x)
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f(x) = 7x^(1/2) − 2,   h(x) = −4x^(1/2) + 1
Find f(x) − g(x) where g(x) = −x^(1/2) + 4 Watch the signs! Distribute the minus!
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⭐⭐
f(x) = 4x²,   g(x) = −3x^(4/3),   h(x) = x^(1/2)
Multiply: f(x) · g(x) — add the exponents!
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⭐⭐
f(x) = 4x²,   h(x) = x^(1/2)
Multiply: f(x) · h(x)
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⭐⭐
f(x) = 4x²,   g(x) = −3x^(4/3)
Divide: f(x) / g(x) — subtract the exponents!
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⭐⭐
f(x) = 4x²,   g(x) = −3x^(4/3),   h(x) = x^(1/2)
Divide: h(x) / f(x)
Part 3 · Inverse Functions UNDO
🔑 "SWAP → SOLVE" Two simple steps to find any inverse:
  Step 1 — SWAP: Replace f(x) with y, then swap x and y.
  Step 2 — SOLVE: Solve for y. That's your f⁻¹(x)!
Remember: f(f⁻¹(x)) = x always. The inverse undoes the function.
✏️ Example — Finding the inverse
f(x) = 3 − 2x
Step 1: y = 3 − 2x → swap → x = 3 − 2y
Step 2: 2y = 3 − x → y = (3 − x)/2 = 3/2 − x/2
f⁻¹(x) = 3/2 − (1/2)x
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Find the inverse: f(x) = (1/5)x + 3
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Find the inverse: f(x) = √(x − 3)
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⭐⭐
Find the inverse: f(x) = √(2x + 5)
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⭐⭐
Find the inverse: f(x) = 4x⁷
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⭐⭐⭐
Find the inverse: f(x) = 4x² + 1  TRICKY!
Hint: This requires a domain restriction (x ≥ 0)
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⭐⭐⭐
Which statement about f(g(x)) and g(f(x)) is ALWAYS true when g = f⁻¹?
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⭐⭐⭐
If f(x) = 3 − 2x, verify by computing f(f⁻¹(4)). What should you get?
★ Super Concept: f(f⁻¹(x)) must always equal x!
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