Algebra and Functions
24th January 2017
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• Algebra and Functions
• Question Focused Approach
• 2 hours - 10mins break @7pm
• Any questions? – Just Ask!!
Write algebraic expressions for the following:
• Multiply 𝑛 by three, then add four.
• Divide 𝑛 by two, then add six.
• Add three to 𝑛, then multiply by two.
• Subtract n from the square of five.
Write algebraic expressions for the following:
• Multiply 𝑛 by three, then add four.
• Divide 𝑛 by two, then add six.
• Add three to 𝑛, then multiply by two.
• Subtract n from the square of five.
Brackets
Exponents
Divide
Multiply
Add
Subtract
Factorisation of Algebraic Expressions
Taking out the highest common factor
4 𝑥 28 𝑥𝑦 = 4𝑥(𝑥2𝑦)
Highest common factor = 4x
2𝑎2𝑏4𝑎𝑏2 + 12𝑎𝑏𝑐 = 2𝑎𝑏 𝑎2𝑏 + 6𝑐
Highest common factor = 2ab
Factorisation by grouping terms
1.
a𝑥 + 𝑎𝑦 + 2𝑥 + 2𝑦
a 𝑥+𝑦 +2 𝑥+ 𝑦
𝑎+2 𝑥+ 𝑦
Check: 𝑥 𝑎 + 2 + 𝑦 𝑎 + 2 = a𝑥 + 𝑎𝑦 + 2𝑥 + 2𝑦
2.
a𝑏 + 5 + 5𝑎 + 𝑏
a𝑏 + 5𝑎 + 𝑎 + 𝑏
a𝑏 + 𝑏 + 𝑎 + 5𝑎
𝑏(𝑎 + 1) + 5(1 + 𝑎)
(𝑏 + 5) (𝑎 + 1)
Difference of two squares
(a + 𝑏)(𝑎 − 𝑏)
a 𝑎−𝑏 +𝑏 𝑎−𝑏
𝑎2 − 𝑎𝑏 + 𝑎𝑏 − 𝑏 2
𝑎2 − 𝑏 2
(a + 𝑏)(𝑎 − 𝑏)
Difference of two squares
1. 𝑥 2 − 9
2. 4𝑦 2 − 12
Solution 1
Solution 2
(𝑥)2 − 3 2
(𝑥 + 3) 𝑥 − 3
(2𝑦)2 − 1 2
(2𝑦 + 1)(2𝑦 − 1)2
Factorising quadratic equations
2
𝑥 + 7𝑥 + 10
2
𝑥 + 5𝑥 + 2𝑥 + 10
(𝑥 + 2)(𝑥 + 5)
Division and Factors
Simplify
Simplify
𝑥 2 + 4𝑥
3𝑥 + 12
3𝑥2
5𝑦2
3𝑥
5𝑦
=
=
3𝑥 2
5𝑦 2
𝑥(𝑥+ 4)
3(𝑥 +4)
=
5𝑦
3𝑥
𝑥
𝑦
×
=
𝑥
3
Divide:
𝑥 + 3 |𝑥 3 + 6𝑥 2 + 11𝑥 + 6
Long Division
𝑥 2 + 3𝑥 + 2
𝑥 + 3 𝑥 3 + 6𝑥 2 + 11𝑥 + 6
𝑥 3 + 3𝑥 2
3𝑥 2 + 11𝑥
3𝑥 2 + 9𝑥
2𝑥 + 6
2𝑥 + 6
Solving equations
Solve the following equations:
(i)
3
𝑥−1
(ii)
2
𝑥−2
-
+
2
𝑥+1
3
𝑥
=1
5
=
𝑥−4
Question (i) – Solution
Solve the equation:
3
𝑥−1
2
𝑥+1
-
=1
Multiply across:

3(𝑥−1)(𝑥+1)
𝑥−1
−
2(𝑥−1)(𝑥+1)
𝑥+1
= 1(x-1)(x+1)
3 x + 1 − 2(x − 1) = 1(x2 +x −x −1)
3x + 3- 2x +2 = x2 - 1
1x + 5 = x2 - 1
x2 -x - 6 = 0
Question (i) - Solution
x2 - x - 6 = 0
(
(x
(x
 (x -
)(
)
)(x
)
3)(x
2)
3)(x + 2)
x = 3, x = -2
Check x = 3 by substituting
32 -(3) - 6 = 0; 9 – 3 - 6 = 0…. correct
Check x = -2 by substituting
(-2)2 - (-2) - 6 = 0; 4 +2 - 6 = 0…. correct
Question (ii) – Solution
Solve the equation:
2
𝑥−2
+
3
𝑥
=
5
𝑥−4
Multiply across:

2(𝑥−2)(𝑥)(𝑥−4)
𝑥−2
+
3(𝑥−2)(𝑥)(𝑥−4)
𝑥
=
5(x−2)(x)(x−4)
𝑥−4
2 x − 4 𝑥 + 3(x − 2)(x − 4) = 5(x−2)(x)
2x2 – 8x + 3(x2 − 2x − 4x +8)= 5 x2 – 10x
2x2 – 8x + 3x2− 6x − 12x + 24= 5x2 – 10x
5x2 – 26x + 24= 5x2 – 10x
Question (ii) – Solution
5x2 – 26x + 24= 5x2 – 10x
 – 26x + 24= – 10x
 24 = 16x
 x = 3/2
 Sub in to check…


2
3/2−2
2
1
−( )
2
+
+
3
3/2
5
3
3/2
=
= −5/2
 -4 + 2 = -2
 -2 = -2 therefore correct
5
3/2−4
Rearranging formula
1. Solve 2𝑎 − 3𝑏 = 5𝑐 in terms of a
2. Solve 5 a − 3b = 2c in terms of a
Solution 1
Solution 2
𝑎 = (5𝑐 + 3b)/2
5a − 15b = 2c
a = 2c + 15b /5
Solving inequalities
𝑥 + 3 ≥ 3𝑥 + 1 , x € Z
𝑥 − 3𝑥 ≥ 1 − 3
−2𝑥 ≥ −2
−𝑥 ≥ −1
𝑥≤1
{1 , 0 , -1 , -2 , ….}
Solving inequalities
𝑥 + 3 ≥ 3𝑥 + 1 , x € Z
𝑥 − 3𝑥 ≥ 1 − 3
−2𝑥 ≥ −2
−𝑥 ≥ −1
𝑥≤1
{1 , 0 , -1 , -2 , ….}
Solving quadratic inequalities
𝟐𝒙𝟐 + 𝟑𝒙 − 𝟐𝟎 < 𝟎
2𝑥 2 + 3𝑥 − 20 = 0
2𝑥 − 5 𝑥 + 4 = 0
2𝑥 − 5 = 0 𝑎𝑛𝑑 𝑥 + 4 = 0
5
𝑥 = 𝑎𝑛𝑑 𝑥 = −4
2
Solution – 4 < x < 5/2
Critical values
Solving modulus inequalities
Solve for x,
3𝑥+1
𝑥+1
=2
3𝑥 + 1
=2
𝑥+1
3𝑥 + 1 = 2 (𝑥 + 1)
3𝑥 + 1 = 2 𝑥 + 2
𝑥=1
Solving modulus inequalities
Solve for x,
3𝑥+1
𝑥+1
=2
3𝑥 + 1
= −2
𝑥+1
3𝑥 + 1 = −2 (𝑥 + 1)
3𝑥 + 1 = −2 𝑥 − 2
3
𝑥=−
5
𝟑
Solution is 𝒙 = 𝟏 and 𝒙 = −
𝟓
Solving inequalities
Solve the following inequalities, x ∈ R:
a.
|2x-1| ≥ 7
b.
|3x+4| ≤ |x+2|
c.
2|x-1| ≤ |x+3|
Solving inequalities
a) Solution
|2x-1| ≥ 7
Step 1: Solve |2x-1| = 7
 (|2x-1|) 2 = 72
 4x2 - 4x + 1= 49
 4x2 - 4x - 48 = 0
 x2 – x – 12 = 0 (now your equation is in the form ax2 + bx + c)
 (x – 4)(x + 3) = 0
x – 4 = 0 ⟶ x = 4
 x + 3 = 0 ⟶ x = -3
Solving inequalities
Step 2:
a=1
Since a >0, use a U shaped graph
-6
-5
-4
-3
-2
5
4
3
2
1
0
-1-1 0
-2
-3
-4
-5
1
2
3
4
5
6
Solving inequalities
Step 3:
x2 – x – 12 ≥ 0 when the curve is above the x axis
 x ≤ -3
x≥4
Solving inequalities
b) Solution
|3x+4| ≤ |x+2|
Step 1: Solve |3x+4| = |x+2|
(|3x + 4|) 2 = (|x + 2|) 2
9x 2 + 24x + 16 = x 2 + 4x + 4
8x 2 +20x + 12 = 0
2x 2 + 5x + 3 = 0 (now your equation is in the form ax2 + bx + c)
(2x + 3)(x + 1) = 0
(2x + 3) = 0 ⟶ x = -3/2
(x + 1) = 0 ⟶ x = -1
Solving inequalities
Step 2:
a=2
Since a > 0, use a U shaped graph
2
1
0
-2
-1
0
-1
-2
1
Solving inequalities
Step 3:
2x 2 + 5x + 3 ≤ 0 when the curve is below the x axis
-3/2 ≤ x ≤ -1
Solving inequalities
c) Solution
2|x-1| ≤ |x+3|
Step 1: Solve 2|x-1| = |x+3|
 (2|x-1|)2 = (|x+3|) 2
 4(x 2 – 2x + 1) = x 2 + 6x + 9
 4x 2 – 8x + 4 = x 2 + 6x + 9
 3x 2 – 14x - 5 = 0 (now your equation is in the form ax2 + bx + c)
 (3x + 1)(x – 5) = 0
 3x + 1 = 0
⟶ x = -1/3
x – 5 = 0
⟶ x=5
Solving inequalities
Step 2:
a=3
Since a > 0, use a U shaped graph
2
1
0
-2
-1
0
-1
-2
1
2
3
4
5
6
7
Solving inequalities
Step 3:
3x 2 – 14x - 5 = 0 ≤ 0 when the curve is below the x axis
-1/3 ≤ x ≤ 5
Functions
• A function is a way of matching the members of a set “A” to a set
“B”.
• A General Function points from each member of "A" to a member of
"B".
• Never has one “A” pointing to more than one B
• More than one “A” can point to the same “B”
Injective
• Injective means that every member of "A" has its own
unique matching member in "B".
• We can have a "B" without a matching "A"
• Is f(x) = x2 where x 𝜖 ℝ injective?
• Is f(x) = x2 where x 𝜖 ℕ injective?
Surjective
• Surjective means that every "B" has at least one matching "A"
There won't be a "B" left out.
• Is f(x) = x2 where x 𝜖 ℝ surjective?
• Is f(x) = x2 where x 𝜖 ℕ surjective?
• Bijective means that a function is both injective and surjective.
Inverse of a Function
• Take 2 functions f(x) and g(x)
• If f(g(x)) = g(f(x)) = x then g(x) is the inverse of f(x)
OR
• g(x) = f-1(x)
𝑥
3
• Take f(x) = 3x – 2 and g(x) = +
2
3
• Show g(x) = f-1(x) = x
• Firstly need to assume that the function is one-toone (injective)
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Finding the inverse of a function
• To find the inverse:
 Replace f(x) with y
 Replace every x with a y and y with an x
 Solve this equation for y
 Replace y with f-1(x)
 Verify that f(f-1(x)) = x and f-1(f(x)) = x
• Find the inverse of f(x) = 3x-2
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Sketching the inverse of a function (i)
• The inverse of a function can actually be
sketched without knowing the inverse.
• Take f(x) = 3x-2
• Firstly, graph some points of the original
function:
43
Sketching the inverse of a function (ii)
• If you know that f(1) = 2 then f-1(2) = 1
• If you are asked to graph a function and its inverse, all
you have to do is graph the function and then switch
all x and y values in each point to graph the inverse.
• Therefore:
• The graph of the inverse is a reflection of the actual
function about the line y = x. This will always be the
case with the graphs of a function and its inverse.
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Question
• Given g(x) = (𝑥 − 3), sketch g(x) and g-1(x) where x
is between 1 and 6
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Exponential Functions
• Exponential functions are functions where f(x) = ax
• An exponential function will never have a negative result.
If f(x) = 3x , can we verify that f(-5) is > 0 and f(5) > 0?
47
Solution
• f(-5) = 3-5 = 0.004
• f(5) = 35 = 243
48
Exponential Functions (ii)
• When the base e is used, the exponential
function becomes f(x) = ex
• Can everyone find e1 on their calculator?
49
Exponential Functions (ii)
• When the base e is used, the exponential
function becomes f(x) = ex
• Can everyone find e1 on their calculator?
• e1 = 2.718
50
Question: 2016 Paper 1
2
𝑒𝑥
• If f(x) = and g(x) = 𝑒 𝑥 - 1 where x 𝜖 ℝ complete
the table below, rounding your answers to two
decimal places.
• Draw graphs of f(x) and g(x) in the domain 0 ≤ x ≤
ln(4)
51
Question
• Use your graphs to estimate the value of f(x) =
g(x)
• Solve f(x) = g(x) using algebra
52
Solution
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Solution
• f(x) = g(x) where x
0.7
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Logarithmic Graphs
• Logarithmic functions are the inverse of exponential functions.
• If b > 0, b ≠ 1 and x > 0 then y = log 𝑏 𝑥 is equivalent to by = x.
• When b = e, loge(x) is often written as ln(x)
• On your calculator there should be a log button and a ln button.
The log button has a base of 10 i.e. b = 10
58
Logarithmic Graphs
• Calculate:
 ln(1)
 ln(0)
 ln(-1)
 ln(e1)
59
Logarithmic Graphs
• Calculate:
 ln(1) = 0
 ln(0) – does not exist
 ln(-1) – does not exist
 ln(e1) = 1
60
Question
• If f(x) = log10(x) and g(x) = loge(x)where
x 𝜖 ℝ complete the table below, rounding your
answers to two decimal places.
• Draw graphs of f(x) and g(x) in the domain 0.5 ≤ x
≤4
61
Solution
• If f(x) = log10(x) and g(x) = loge(x)where
x 𝜖 ℝ complete the table below, rounding your
answers to two decimal places.
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Solution
• Draw graphs of f(x) and g(x) in the domain 0.5 ≤ x
≤4
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Useful Resource for Project Maths
• www.themathstutor.ie
• Normally €49.99
• Discount code: SOA 2017
• €29.99 for the year
• Access only to Sept 2017
Next lecture:
• https://web.actuaries.ie/students/tutorials
• 31 January 2017
• Same location
• 6 - 8 pm
• Calculus
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