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maths1-week5
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Deep study for Quiz 2 week 5. Master composition order first, then invertibility, then exponential growth rules.
Week map
Vertical/horizontal line tests → composition (f∘g) → inverse f−1 → exponential ax → domain discipline throughout.
Composition notation
- (f∘g)(x) → “f composed with g at x” → apply g first, then f → (f∘g)(x)=f(g(x)).
- f(g(x)) → inner function output becomes outer input → parentheses matter when substituting.
- Domain of composition → all x where g(x) is in domain of f.
Mini-example: f(x)=2x+1, g(x)=x2. Then (f∘g)(3)=f(g(3))=f(9)=19. But (g∘f)(3)=g(f(3))=g(7)=49 — order changes the answer.
Vertical and horizontal line tests
- Vertical line test: graph is a function iff no vertical line hits twice.
- Horizontal line test: function is injective (one-to-one) iff no horizontal line hits twice — needed for invertible function on its domain.
Inverse notation
- f−1(y) → input that f sends to y → not the same as 1/f(x).
- Inverse swaps roles: if f(a)=b then f−1(b)=a.
- Domain of f−1 = range of f; range of f−1 = domain of f.
- Graph of f−1 reflects graph of f over line y=x.
Mini-example: f(x)=3x−2 on R. Swap: x=3y−2, solve y=3x+2. So f−1(x)=3x+2. Check: f−1(5)=7/3 and f(7/3)=5.
Exponential notation
- y=ax → exponential with base a → a>0, a=1.
- a0=1 for any valid base; am+n=am⋅an; (am)n=amn.
- ax>0 always — range is (0,∞).
Mini-example: 23=8; 2−1=1/2; (1/2)2=1/4. Base between 0 and 1 gives decay; base >1 gives growth.
Trap: xa (power function) vs ax (exponential) — variable position differs.
Pattern families
Easy — Evaluate composition
Substitute entire inner expression into outer. Simplify stepwise. Watch signs when replacing x inside f.
Medium — Find inverse of linear or simple rational
Swap x and y, solve for y. State domain of inverse explicitly. Verify f(f−1(x))=x on valid domain.
Hard — Invertibility and restricted domains
Use horizontal line test or monotonicity. Square root and squaring need domain restriction (x≥0) for bijection. Composition with inverse simplifies to identity only on matched domains.
Worked mini-examples
Example 1 — Composition order.
f(x)=x+4, g(x)=2x. (f∘g)(5)=f(10)=14. (g∘f)(5)=g(9)=18.
Example 2 — Nested formula.
f(x)=x2+1, g(x)=x−3. (f∘g)(x)=(x−3)2+1=x2−6x+10.
Example 3 — Inverse of linear.
f(x)=2x−1. Swap: x=2y−1 → 2x=y−1 → f−1(x)=2x+1.
Example 4 — Exponential evaluation.
32=9; 3−2=1/9; 31/2=3.
Example 5 — Not invertible without restriction.
f(x)=x2 on all R fails horizontal line test (f(2)=f(−2)=4). Restrict to x≥0 to define inverse f−1(x)=x.
Traps
- Writing (f∘g)(x) as f(x)⋅g(x).
- Treating f−1 as 1/f(x).
- Taking domain of f−1 equal to domain of f.
- Applying outer function before inner in composition.
- Confusing exponential base with exponent: 2x vs x2.
Diagnostic (try yourself)
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If f(x)=3x−1 and g(x)=x2, find (g∘f)(2) and (f∘g)(2).
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Find f−1(x) for f(x)=5x+4. What is the domain of f−1 if domain of f is R?
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Is f(x)=∣x∣ injective on R? What restriction makes it bijective from R to [0,∞)?
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Simplify (f∘g)(x) when f(x)=x and g(x)=x+5. State the domain of the composition.
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Evaluate 23 and 32. Which is larger? What is 2−3?