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f(x) = x² | f→x² | f = {(−1,1),(0,0),(1,1),(2,4)} |
---|---|---|
f of x equals x- squared | f maps x to x² | Ex : if 2 is input, 4 is output |
Examples:
f: N→N : x→2x
f: R→R : x→x²
h∘g(x) or (h∘g)(x) or h(g(x)) or hg(x)
ie · f·g(x) ≠ g·f(x)
y = a(x-h)²+k
a>0 U shaped a<0 ∩ shaped
vertex h, k axis of summetry x = h
Put ax² + bx + c in the form a(x² + x + )
Halve the coefficient of x and put it in bracets ...a[(x + ())² +...
Subtract half the coefficient of x squared and include
a[(x + ())² - (())² + ]
Evaluate and simplify then multiply by a
A cubic function - f in x involves an x³ term and is of the form f(x) = ax³ + bx² + cx + d where a, b, c and d are constants (a≠0) and x is a variable
If coefficient of x³ is positive - graph starts low, ends high
If coefficient of x³ is negative - graph starts high, ends low
Note: Graph of function f(x)=abˣ will pass through point (0,a) (At x=0, y=ab⁰ =a(1) = a )
Graph of exponential function will never touch or cross horizontal axis
If exponent is x and the base is greater than 1, curve slopes upwards
Transformations: y = a log (x±b)±c
Multiplying by a does not affect roots only compresses or elongates function. (Ex multiplying by 0.5 = all y values halved) (compared to original)
-b - shift b units to right ● +b - shift b units to left
+c - shift c units up ● -c - shift c units down
Horizontal line test for Injectivity- If the horizontal axis, is the input axis, then, if any horizontal line cuts the graph of function at no more than one point function is injective
Test by algebra - Take values of ∈R a, b and let f(a) = f(b)
Surjective function (onto) - a function which every element in the codomain has at least one matching element in the domain.
Codomain = Range
The function f is surjective if ∀ b∈B, ∃ a ∈A such that f(a)=b (∃ "there exists")
Horizontal Line test for surjectivity - if f is surjective then every horizontal line y=b where b ∈ B, intersects graph of f at least one point.
Algebraically -
Bijective function - a function f is said to be bijective if it is both injective and surjective
To change a function to be bijective:
If two functions f and g are defined so that f: A→ B and g: B→ A, then, if (f·g) (x) = (g·f) (x) = x we say that f and g are inverse functions of each other
A function f has an inverse function f⁻¹ if and only if f is bijective
A function f that is bijective will have a unique inverse function f⁻¹
Steps to find inverse with example
1. Let f(x) = y 1. y = 2x+3 , x ∈ R
2. Express x in terms of y 2. y-3 = 2x ⇒ x = y-3/2
3. Rewrite as f-1(x) replacing 3. f-1(x) = x-3/2 , x ∈ R y with x
For any invertible function f and its inverse f-1, we get the following: If (c,d) is a point on graph of f, then (d,c) is point on graph of f-1 point on graph of f-1 graphs of f and f-1 are reflections of each other about line y = x
Remember: cos𝜃 = x - coordinate sin𝜃 = y - coordinate
Period - the horizontal distance required for the graph to complete one cycle
Range - the set of all actual outputs of the function
Amplitude - the number of units the graph goes up or down from the midline of the graph
Frequency - Number of waves that pass fixed point in a given amount of time
Period: 2π Range: [-1,1]
Period: 2π Range: [-1,1]
Period: π Range: (-∞, ∞)
To Note: If a, n ∈ N then :
ie : Sin𝜃 and Cos c𝜃 Period = have
|b| = amplitude (becomes inverted if b is negative) → = period
a + bsin𝜃
midway line → range = [a-b, a+b] or [a+b, a-b]
Amplitude → period()
a + bcos𝜃
midway line → range
Ex:
This is the graph of sin x shifted to the right
Graph of y = sin⁻¹x
Graph of y = cos⁻¹x
The inverse function g⁻¹(x) = tan⁻¹x
• The ranges of the inverse trigonometric functions are known as principal values. When we use calculator to find the inverse trigonometric functions the answer will always be within the range of principal values
Trig Ratio | Principal Values |
---|---|
sin⁻¹ | -90° ≤ θ ≤ 90° or -π/2 ≤ θ ≤ π/2 |
cos⁻¹ | 0° ≤ θ ≤180° or 0 ≤ θ ≤ π |
tan⁻¹ | -90° < θ < 90° or -π/2 ≤ θ ≤ π/2 |
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