POLI270 - Day 2
Fundamentals (notation, IV, degree, increasing, decreasing, extrema, domain, slope, functional form)
Limits (piece wise, asymptotes, instantaneous rate of change, limit process for
derivative)
Function - rule that assigns a single output value to every input. The output is
the dependent variable and endogenous. The output is independent and exogeneous.
Types: Monomials: f (x) = axk , g(x), h(x). a is the coefficient, k is the degree.
Polynomials: addition/subtraction of monomials
Look at general graphs, x intercept; y-intercept; vertical line test; finding zeroes.
The degree is the highest monomial.
Example
1a) f (x) = x2 − 4
1b) f (x) = x3 + x2 − 2x
(x)
Rotational function: ratio of polynomials fq(x)
Domain: set of numbers in x at which function is defined.
Range: set of values f (x) is assigned by.
√
Undefined by mathematical properties of q(x) = 0 for mathematical reasons −
Interval Notation:
•◦
[]()
inclusive exclusive
closed open
x > a, x < a
x ≥ a, x ≤ a
Example:
3
2a) x3 ∪ x33
√
x2 −4
√
2b) x2x+2
−4 versus x+2
Composite functions - taking the function of a function
- applying a function to a function of x
f (x) = g(h(x)) = (g ◦ h)(x)
Example
3a) f (x) = 2x + 5, g(x) = 3x2 , f (g(x)), g(f (x))
3b) f (x) = 4x2 − 1, g(x) = 3x + 2, f (g(x)), g(f (x)), g(g(x))
Piecewise function - rule determining a functions value varies over its domain.
Example
1
x, 0 ≤ x ≤ 1
f (x) =
(1)
2 − x, 1 < x ≤ 2
x, x < 0
g(x) =
(2)
x2 , x > 0
x − 1, x < 0
√
0, x ≤ 0
h(x) =
(3)
1
Slope - change in y with a corresponding one unit change of x y2 −y
1 . How
much f (x) increases for each unit increased in y.
Linear functions have the same slope throughout
Applying slope formula to non-linear functions returns the average change
∆y
y2 − y1
=
∆x
x2 − x1
(4)
19 − 1
18
∆y
=
=
=3
∆x
6−0
6
∆y
9−1
8
=
= =2
∆x
4−0
4
∆y
3−1
2
=
= =1
∆x
2−0
2
(5)
∆y
7−9
3
=
=
= −3
∆x
7−0
−1
∆y
6−4
=
=2
∆x
4−3
∆y
6−2
=
=2
∆x
4−2
(6)
∆y
4−1
3
=
= =1
∆x
1 − −2
3
(7)
f (x) = mx + b
Graph 1:
f (x) 12 x2 + 1
Graph 2:
Graph 3:
Differentiation Rules
2
Power Rule
f (x) = xk
(8)
f 0 (x) = kxk−1
Coefficient
h(x) = axk
(9)
h0 (x) = akxk−1
Constant
g(x) = a
(10)
g 0 (x) = 0
Addition/Subtraction
h(x) = f (x) ± g(x)
(11)
h0 (x) = f 0 (x) ± g 0 (x)
Product
h(x) = f (x)g(x)
h0 (x) = f (x)g 0 (x) + f 0 (x)g(x)
(12)
Quotient
f (x)
g(x)
0
g(x)f (x) − f (x)g 0 (x)
h0 (x) =
[g(x)]2
h(x) =
√
Constant: 8, − 12 , 3
Coefficient: 8x2 , −x3 , 2x1 3
√
Product Rule: (x2 + 1)(x3 + 3), x( x22 − 1)
√
√
3
Power Rule: x2 , x1/2 , x5 , x2
4
Addition/Subtraction: x + 12x, √1x + 3x, 4x2 − 5 −
Quotient Rule:
2
(13)
1
x
2
x −1
x +3
√
x2 +1 , (x+1) x
Application of Derivatives
Increase: f 0 (x) > 0 open interval of increasing x1 > x2 , f (x1 ) > f (x2 )
Decrease: f 0 (x) < 0 open interval of decreasing x1 > x2 , f (x1 ) < f (x2 )
lim
h→0
f (x0 + h) − f (x0 )
= f 0 (x0 ) > 0
h
3
(14)
h is small and positive, thus f (x1 ) > f (x2 ) candidates for local extrema
Critical points occur where f 0 (x) = 0 or undefined.
1st derivative test finds points within interval.
Absolute extrema need to include endpoints.
Concavity:
f 00 (x) > 0 concave up secant above derive up f 0 (x) = 0 or undefined. Potential
POI
f 00 (x) < 0 concave down secant below derive down.
2nd derivative test for x0 as critical point if f 00 (x) > 0x0 is a relative max
If f 00 (x) < 0, x0 is a relative min
Chain Rule + Composite Functions
d
(h(g(x)) = h0 (g(x)) − g 0 (x)
dx
dy
dy du
=
dx
du dx
y = (3x2 + 1)3
g(x) = 3x2 + 1, h(x) = x3
2(3x2 + 1)2 (6x)
u = 3x2 + 1
y = u3
3u2 6x
3(3x1 + 1)2 6x
p
y = x2 − 4
√
g(x) = x2 − 4, h(x) = x
1 2
(x − 4)−1/2 (2x)
2 √
y = u, u = x2 − 4
1 −1/2
u
2x
2
1 2
(x − 4)−1/2 (2x)
2
Exponentials:
4
(15)
2x can never be negative
ar ∗ as = ar+s
1
a−r = r
a
ar
= ar−s
as
(ar )s = ar∗s
a0 = 1
x 0
(e ) = ex
(eu(x) )0 = eu(x) ∗ u0 (x)
1
e = lim (1 + )n e4 ∗ e−2 = e2
n→∞
n
1
e4
e4 ∗ 2 = 2 = e2
e
e
(e2 )3 = e6
y = e2x , y 0 = e2x ∗ 2
2
2
y = e4x , y 0 = e4x ∗ 8x
5
(16)
Logarithms
z = ax
x = loga z
log(r ∗ s) = log(r) + log(s)
log(1) = 0
1
log( ) = −log(s)
s
r
log( ) = log(r) − log(s)
s
log(rs ) = s ∗ log(r)
loge = ln
1
ln(x)0 =
x
u0 (x)
0
ln(u(x)) =
u(x)
ln(12) = ln(2) + ln(6) = ln(3) + ln(4)
5x
ln( ) = ln(5) + ln(x) − ln(2)
2
ln(x2 ) = 2ln(x) = ln(x) + ln(x)
(17)
1
ln( ) = ln(8)−1 = −ln(8) = −ln(23 ) = −3ln(2)
8
2x
y = ln(x2 + 3), y 0 = 2
x +3
5
0
y = ln(3x) − 6, y =
5x − 6
Log Conversion
62 = 36
log6 (36) = 2
(18)
91/2 = 3
1
log9 (3) =
2
Use existing graphs for critical points
increasing decreasing
Example 1) f (x) = x3 − 2x to find max min(first derivative test)
2)
f (x) = 2x6 − 3x4 + 2
f 0 (x) = 12x5 − 12x3 = 12x3 (x2 − 1) = 12x3 (x + 1)(x − 1)
6
(19)
3)
x3
(20)
4)
x4/3 − 4x1/3
4x−2/3 (x − 1)
4x − 1
x2/3
(21)
5)
f (x) = −x3 + 12x + 5 for − 3 ≤ x ≤ 5
(22)
x2 − 4x + 4 for 1 ≤ x < ∞
(23)
6)
2nd derivative - what is happening to the derivative. What will happen if the
derivative is increasing.
2nd derivative test example
y = x4 − 4x3 + 10
y 0 = 4x3 − 12x2 = 4x2 (x − 3); x = 0, 3
00
(24)
2
y = 12x − 24x = 12x(x − 2); x = 0, 2
x
x2 + 1
x2 + 1 − 2x2
−x2 + 1
(x2 + 1)(1) − x(2x)
f 0 (x) =
=
= 2
2
2
2
2
(x + 1)
(x + 1)
(x + 1)2
(x2 + 1)2 (−2x) − (−x2 + 1)(2)(x2 + 1)(2x)
f 00 (x) =
(x2 + 1)4
f (x) =
7
(25)