y = f(x) or f : x ↦ y.f(x)=2x+1 on ℝ.f(x)=x² on ℝ.f⁻¹. Example: f(x)=3x‑4 ⇒ f⁻¹(x)= (x+4)/3.| Function type | Typical domain | Typical range |
|---|---|---|
Linear y=mx+c | ℝ | ℝ |
Quadratic y=ax²+bx+c | ℝ | y ≥ k if a>0; y ≤ k if a<0 |
Rational y = p(x)/q(x) | q(x) ≠ 0 | Depends on asymptotes |
Trigonometric sin x, cos x | ℝ | [-1, 1] |
Exponential y = aˣ (a>0, a≠1) | ℝ | (0, ∞) |
Logarithmic y = logₐx (a>0, a≠1) | (0, ∞) | ℝ |
sin x and cos x are many‑one on ℝ because many angles have the same sine or cosine value.sin x is one‑to‑one on [-π/2, π/2] → inverse sin⁻¹x (arcsin) defined on [-1, 1].cos x is one‑to‑one on [0, π] → inverse cos⁻¹x (arccos) defined on [-1, 1].When a function is not one‑to‑one on its natural domain, we restrict the domain so that an inverse exists.
Example – square‑root and square:
f(x)=√x, domain = [0, ∞).f⁻¹(x)=x² but only if we restrict the inverse’s domain to x ≥ 0; otherwise x² would not be one‑to‑one.y=√x in the line y=x.Given f(x) and g(x):
fg(x) = f(g(x)) and gf(x) = g(f(x)).(f∘g)(x) = f(g(x)).Example:
f(x) = √x, g(x) = x² + 1
fg(x) = f(g(x)) = √(x²+1) (gf(x) = g(f(x)) = (√x)² + 1 = x + 1, domain x ≥ 0)
To obtain the graph of f⁻¹:
f on a set of axes.y = x.f across the line y = x; the reflected curve is the graph of f⁻¹.f(x) = ax² + bx + c (a ≠ 0)f(x) = a (x – h)² + k where the vertex is (h, k).f(x)=ax²+bx+c.a from the quadratic terms: f(x)=a[x² + (b/a)x] + c.(b/2a)² inside the brackets:f(x)=a[x² + (b/a)x + (b/2a)² – (b/2a)²] + c
f(x)=a[(x + b/2a)² – (b/2a)²] + c
a and combine constants:f(x)=a(x + b/2a)² + c – b²/4a
h = –b/2a, k = c – b²/4a
For f(x)=ax²+bx+c:
f'(x)=2ax + b.f'(x)=0 ⇒ x = –b/2a.x back into f(x) to obtain k. The sign of a tells you whether the stationary point is a minimum (a>0) or a maximum (a<0).For the quadratic equation ax² + bx + c = 0 the discriminant is Δ = b² – 4ac:
Δ > 0 → two distinct real roots.Δ = 0 → one repeated real root (the vertex lies on the x‑axis).Δ < 0 → no real roots (the parabola does not intersect the x‑axis).(h, k) (by completing the square or differentiation).x = h.x = 0 → value c.x = [–b ± √Δ]/(2a).a>0 (minimum at k) or downwards if a<0 (maximum at k).To solve ax² + bx + c < 0 (or ≤, >, ≥):
r₁, r₂ (if any). If Δ < 0, the sign of the quadratic is the sign of a for all x.a>0 the parabola is below the x‑axis between the roots.a<0 it is above the x‑axis between the roots.Find the maximum value of f(x)=‑3x² + 12x – 5.
‑3: f(x)=‑3(x² – 4x) – 5.‑3[(x‑2)² – 4] – 5.‑3(x‑2)² + 12 – 5 = ‑3(x‑2)² + 7.(h, k) = (2, 7). Since a = ‑3 < 0, k = 7 is the maximum.Find the minimum of g(x)=4x² – 8x + 3.
g'(x)=8x – 8.8x – 8 = 0 ⇒ x = 1.g(1)=4·1² – 8·1 + 3 = –1.a = 4 > 0, the point (1, ‑1) is a minimum.Determine the nature of the roots of 2x² + 3x – 2 = 0.
Δ = 3² – 4·2·(‑2) = 9 + 16 = 25 > 0 → two distinct real roots
x = [‑3 ± √25]/(2·2) = (‑3 ± 5)/4 ⇒ x = –2 or x = ½.
Solve x² – 5x + 6 ≥ 0.
(x‑2)(x‑3) ≥ 0.x = 2, 3.a = 1) gives non‑negative values for x ≤ 2 or x ≥ 3.If a polynomial P(x) is divided by (x‑a), the remainder is P(a).
Example: P(x)=2x³ – 5x² + x – 7. Remainder on division by (x‑2) is P(2)=2·8 – 5·4 + 2 – 7 = –9.
(x‑a) is a factor of P(x) iff P(a)=0.
Example: Show that (x+1) is a factor of x³ + 2x² – x – 2.
P(‑1)= (‑1)³ + 2(‑1)² – (‑1) – 2 = 0 ⇒ (x+1) is a factor.
Worked example: Factor f(x)=x³ – 6x² + 11x – 6.
f(1)=0 ⇒ (x‑1) is a factor.f(x) = (x‑1)(x² – 5x + 6).(x‑2)(x‑3).f(x) = (x‑1)(x‑2)(x‑3).General forms:
|ax + b| = c → ax + b = c or ax + b = –c (provided c ≥ 0).|ax + b| < c → ‑c < ax + b < c.|ax + b| > c → ax + b < –c or ax + b > c.Example: Solve |2x – 3| ≤ 5.
‑5 ≤ 2x – 3 ≤ 5.‑2 ≤ 2x ≤ 8.‑1 ≤ x ≤ 4.If an equation contains only even powers of x, set u = x².
Example: Solve x⁴ – 5x² + 4 = 0.
u = x² → u² – 5u + 4 = 0.(u‑1)(u‑4)=0 ⇒ u = 1 or u = 4.x² = 1 ⇒ x = ±1; x² = 4 ⇒ x = ±2.A cubic of the form y = a(x‑p)(x‑q)(x‑r) has up to three real roots (the x‑intercepts) and up to two turning points where dy/dx = 0.
Sketching steps:
a (a>0 → left → down, right → up; a<0 opposite).y' = 3ax² + 2bx + c and solving y'=0.Example inequality: Solve x³ – 4x² + x + 6 > 0.
(x+1)(x‑2)(x‑3) > 0.‑1, 2, 3.(‑∞,‑1) negative
(‑1, 2) positive
(2, 3) negative
(3, ∞) positive
(‑1, 2) ∪ (3, ∞).Standard methods: elimination or substitution.
Example: Solve
⎧ 2x + 3y = 7
⎨ 5x – y = 4
⎩
y: multiply the second equation by 3 → 15x – 3y = 12.(2x+3y) + (15x‑3y) = 7 + 12 ⇒ 17x = 19 ⇒ x = 19/17.5(19/17) – y = 4 ⇒ y = 5·19/17 – 4 = 95/17 – 68/17 = 27/17.(x, y) = (19/17, 27/17).Typical strategy:
Example: Solve
⎧ y = x² – 4
⎨ x + y = 5
⎩
y from the first into the second: x + (x² – 4) = 5 ⇒ x² + x – 9 = 0.Δ = 1 + 36 = 37 → two real roots:x = [‑1 ± √37]/2.
y = x² – 4 gives the two ordered pairs.ax²+bx+c: h = –b/(2a), k = c – b²/(4a)x = [–b ± √(b²‑4ac)]/(2a)Δ = b² – 4acf'(x)=2ax+bR = P(a) when P(x) ÷ (x‑a)(x‑a) is a factor ⇔ P(a)=0a, b, c in ax²+bx+c.x = –b/(2a).x into the original expression to obtain the extremum value k.a<0) or a minimum (a>0).f(x)=a(x‑h)²+k.