Download - Functions and Graphs.ppt
![Page 1: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/1.jpg)
Get Startedgoodbyegoodbye
Revision NotesFunctions and Graphs
Higher Maths
![Page 2: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/2.jpg)
Functions and graphs You should know the meaning of the terms domain and range of a function;
f : x → sin (ax + b),
f : x → cos (ax + b)
f: x → ax
(a > 1 and 0 < a < 1, x ∈ R) f: x → logax (a > 1, x > 0)
Recognise the probable form of a function from
its graph
Given the graph of f(x) draw the graphs of related functions, where f(x) is a simple
polynomial or trigonometric function
Polynomial functions functions with restricted domain
Inverse of a functionComposite function
Complete the square.Complete the square.
Radian measure.Radian measure.
![Page 3: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/3.jpg)
In both graphs• a = 360 ÷ wavelength. … the number of waves in 360˚.• b = shift to the left a.
y=sin(ax + b)
-1.5
-1
-0.5
0
0.5
1
1.5
y-sin(ax+b)
y-sin(b)
y=sin(ax)
y=sin(ax + b)
-1.5
-1
-0.5
0
0.5
1
1.5
y-sin(ax+b)
y-sin(b)
y=sin(ax)
Shift to leftShift to left
wavelengthwavelength
y=cos(ax + b)
-1.5
-1
-0.5
0
0.5
1
1.5
y= cos(ax + b)y= cos(b)y = cos(ax)
y=cos(ax + b)
-1.5
-1
-0.5
0
0.5
1
1.5
y= cos(ax + b)y= cos(b)y = cos(ax)
wavelengthwavelength
Shift to leftShift to left
-1.5
-1
-0.5
0
0.5
1
1.5
0 20 40 60 80 100 120
Imagine the red curve, the corresponding function of the form y = sin(ax).
The wavelength is 120˚The shift to the left is 60 – 40 = 20˚
a = 360 ÷ 120 = 3
b = 20 3 = 60The blue curve has equation y = sin(3x + 60)˚
Example: Find the equation of the blue curve if it is of the form y = sin(ax + b).
Test Yourself?
![Page 4: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/4.jpg)
y = a^xy = a^x
(0, 1)(0, 1)
(1, a)(1, a)
; 0 > a > 1; 0 > a > 1y = a^xy = a^x
(0, 1)(0, 1)
(1, a)(1, a)
; a > 1; a > 1
Exponential functions
Logarithmic functions
y = logaxy = logax
(1, 0)(1, 0)
(a, 1)(a, 1)
€
y = 1a( )
x= a−x
Note:
An exponential function is the inverse of the corresponding logarithmic function.
€
loga (ax ) = x
aloga x = x
When a = e = 2·71828…
the function is called the exponential function.
Test Yourself?
![Page 5: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/5.jpg)
polynomial functions 1
-10
-8
-6
-4
-2
0
2
4
6
8
10
-5 -4 -3 -2 -1 0 1 2 3 4 5
constant
linear
quadratic
polynomial functions 1
-10
-8
-6
-4
-2
0
2
4
6
8
10
-5 -4 -3 -2 -1 0 1 2 3 4 5
constant
linear
quadratic
polynomial functions 2
-6
-4
-2
0
2
4
6
8
10
-4 -3 -2 -1 0 1 2 3 4 5
cubic (order 3)
quartic (order 4)
polynomial functions 2
-6
-4
-2
0
2
4
6
8
10
-4 -3 -2 -1 0 1 2 3 4 5
cubic (order 3)
quartic (order 4)
Polynomial functions
In general a polynomial of order n will have at most n real roots and at most (n – 1) stationary points.
e.g. a cubic can have, at most, 3 real roots and 2 s.p.s
If a cubic has 3 real roots a, b and c then its equation will be of the form y = k(x – a)(x – b)(x – c) where k is a constant.
If a, b and c are known then k can be calculated if one point the curve passes through is also known.
When the brackets are expanded the cubic will have the form
y = px3 + qx2 + rx + s, where p, q, r, and s are constants and p ≠ 0
Similar statements can be made of the other polynomials.Test
Yourself?
Constant: y = a
linear: y = ax + b
quadratic: y = ax2 + bx + c
cubic: y = ax3 + bx2 + cx + d
quartic: y = ax4 + bx3 + cx2 + dx + e
![Page 6: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/6.jpg)
Reciprocal functions
y = a/x
-5 -4 -3 -2 -1 0 1 2 3 4 5
y = a/x
-5 -4 -3 -2 -1 0 1 2 3 4 5
Restricted domain
€
y =a
f x( ); f (x) ≠ 0
€
y =3
x + 2 then x ≠ −2
The denominator can never equal zero.
So a value of x which makes this happen is not in the domain.
For example
y = √x
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2 3 4 5 6 7 8 9
y = √x
0
0.5
1
1.5
2
2.5
3
3.5
0 1 2 3 4 5 6 7 8 9
Square root function
The term within the radical sign must always be ≥ 0
So any value of x which makes this negative is not in the domain.
€
y = f (x); f (x) ≥ 0
€
y = x + 3 then x ≥ −3
For example
y = tan(x) then x ≠ 90, 270 or 90 + 180n
where n is an integer.
y = log(x) then x > 0
Others
y = sin–1(x)y = cos–1(x)
then –1 ≥ x > 1
Test Yourself?
![Page 7: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/7.jpg)
inverse
If f(g(x)) = x and g(f(x)) = x for all x in the domain then we say that f is the inverse of g and vice versa.
The inverse of f is denoted by f–1.
Examples [over suitable domains]
• f(x) = x2 … f–1(x)= √x
• f(x) = sin(x) … f–1(x) = sin–1(x)
• f(x) = 2x + 3 … f–1(x) = (x – 3)/2
• f(x) = loga(x) … f–1(x) = ax
• f(x) = ex … f–1(x) = ln(x)
For the Higher exam you need not know how to find the formula for the inverse of any function.
composites
![Page 8: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/8.jpg)
Composite functions
Example
Suppose we have two functions: f(x) = 3x + 4 and g(x) = 2x2 + 1.
We can use these definitions to create new functions:
1 f(f(x))
= f(3x + 4)
= 3(3x + 4) + 4
= 9x + 16
2 g(g(x))
= g(2x2 + 1)
= 2(2x2 + 1)2 + 1
= 8x4 + 8x2 + 3
3 f(g(x))
= f(2x2 + 1)
= 3(2x2 + 1) + 4
= 6x2 + 7
4 g(f(x))
= g(3x + 4)
= 2(3x + 4)2 + 4
= 18x2 + 48x + 36
Things to note:
• A composition can be made from more than two functions
• Considering examples 1 and 2 leads to recurrence relations e.g. f(f(f(f(x)))))
• In general f(g(x)) ≠ g(f(x)) … the order in which you do things are important.
• If either f or g have restrictions on their domain, this will affect the domain of the composite function.
• If f(g(x)) = x for all x in the domain then we say that f is the inverse of g … it can be denoted by g–1
Test Yourself?
![Page 9: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/9.jpg)
Related functions
y = f(x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
y = f(x) + a
y = f(x + a)
y = af(x)
y = f(ax)
y = –f(x)
y = f(–x)
y = f–
1(x)y = f ´(x)
[The inverse]
[The derivative]
[x-translation of –a]
[y-translation of a]
[reflection in x-axis]
[reflection in y-axis]
[stretch in y-direction]
[squash in x-direction]
Test Yourself?
![Page 10: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/10.jpg)
Completing the square
€
x + a( )2
= x 2 + 2ax + a2
⇔ x 2 + 2ax = x + a( )2
− a2
We can use this identity to simplify quadratic expressions.
Example 1
Express x2 + 6x + 1 in the form (x + a)2 + b
Given x2 + 6x + 1
By inspection a = 6 ÷ 2 = 3
So x2 + 6x + 1 = (x + 3)2 – 32 + 1
= (x + 3)2 – 8
Note: a = 3 and b = –8
Example 2(a) Express 3x2 + 12x + 1 in the form a(x +
b)2 – c(b) Find the smallest value the expression
can take.
(a) Given 3x2 + 12x + 1, Take 3 out as a common factor leaving the
coefficient of x2 as 1So 3(x2 + 4x) +1 … focus on the red text.
By inspection a = 4 ÷ 2 = 2So we get 3(x2 + 4x) +1 = 3[(x + 2)2 – 22] +
1= 3(x + 2)2 – 12 + 1= 3(x +2)2 – 11
• The smallest a perfect square can be is zero.So the smallest the expression can be is 0 – 11 = 11.Ths happens when x = –2.
Test Yourself?
![Page 11: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/11.jpg)
radians
We can measure angle size using the degree (90˚ - 1 right angle)
We can measure angle size using the grad (100 grads - 1 right angle)
1 radian R
R
R
Mathematicians find it convenient to use the radian.(π/2 radians = 1 right angle.
degreedegree π π ÷ 180÷ 180 radianradian
degreedegree÷ π÷ π 180 180radianradian
The values are often given in terms of π.
€
30° = π6 radians
60° = π3 radians
90° = π2 radians
45° = π4 radians
180° = π radians
Test Yourself?
![Page 12: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/12.jpg)
[x-translation of –a]
y = f(x + a)
y = f(x+1)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
–1
![Page 13: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/13.jpg)
y = f(x) + a[y-translation
of a]
y = f(x)-5
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3–5
![Page 14: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/14.jpg)
y = –f(x)[reflection in
x-axis]
y = –f(x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
![Page 15: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/15.jpg)
y = f(–x)[reflection in
y-axis]
y = f(–x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
![Page 16: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/16.jpg)
y = f ´(x)
[The derivative]
y = f'(x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
+
+
+
+
+
+
+
+
–
–
–
–
Gradient of the function is shown in red
Stationary points of the function correspond to zeros of the derived function.
Positive gradients of the function correspond to Parts below the axis on the derived function.
Negatve gradients of the function correspond to Parts above the axis on the derived function.
![Page 17: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/17.jpg)
y = f–
1(x)[The inverse]
Inverse of a function
0
1
2
3
4
5
6
7
0 1 2 3 4 5 6 7
function
y=x
inverse
When a function has an inverse then, if (x, y) lies on the graph of the function,
(y, x) lies on the graph of the inverse function.
…one is the reflection of the other in the line y = x.
inverse of a function
-5
0
5
10
15
-5 0 5 10 15
function
y=x
inverse
Note that the example function does not have an inverse. The reflection in y = x has, for example, 3 values corresponding to x = 5.
![Page 18: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/18.jpg)
y = f(ax)[squash in x-
direction]
y = f(2x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
![Page 19: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/19.jpg)
y = af(x)[stretch in y-
direction]
y = 3f(x)
-15
-10
-5
0
5
10
15
-4 -3 -2 -1 0 1 2 3
![Page 20: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/20.jpg)
![Page 21: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/21.jpg)
Using suitable units, the distance of the tip of the rotorto the tail of a helicopter can be calculated using a formula of the formD = 3sin(ax + b) + 10. The graph is shown below.
What are the values of a and b
Rotor Distance
0123456789
1011121314
0 15 30 45 60 75 90 105 120
revealreveal
![Page 22: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/22.jpg)
Using suitable units, the distance of the tip of the rotorto the tail of a helicopter can be calculated using a formula of the formD = 3sin(ax + b) + 10. The graph is shown below.
What are the values of a and b
Rotor Distance
0123456789
1011121314
0 15 30 45 60 75 90 105 120
Note:
The 10 translates the sine wave 10 units up.
The 3 stretches the wave by a factor of 3 in the y-direction.
The wavelength, by inspection, is 120˚.
a = 360 ÷ 120 = 3
One would expect the first peak of y = sin(3x) to occur at 90 ÷ 3 = 30.
It occurs at 15. Thus the shift to the left is 30 – 15 = 15.
So b = shift a = 15 3 = 45.
The equation is:
D = 3sin(3x + 45) + 10
![Page 23: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/23.jpg)
Eiffel Tower
0
50
100
150
200
250
300
350
0 10 20 30 40 50 60 70
width from centre (m)
height (m)
The profile of the Eiffel tower can be modelled by the formula
y = a ln(bx)
where a and b are constants and y m is the height of a spot on the profile and x is its distance measured horizontally from the centre.
When y = 0, x = 63. When y = 42, x = 40.
Find the values of a and b.
revealreveal
![Page 24: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/24.jpg)
Eiffel Tower
0
50
100
150
200
250
300
350
0 10 20 30 40 50 60 70
width from centre (m)
height (m)
The profile of the Eiffel tower can be modelled by the formula
y = a ln(bx)
where a and b are constants and y m is the height of a spot on the profile and x is its distance measured horizontally from the centre.
When y = 0, x = 63. When y = 42, x = 40.
Find the values of a and b.
When y = 0, x = 63
y = a ln(bx) 0 = a ln (63b) ln(63b) = 0 63b = 1 b = 1/63
When y = 42, x = 40 y = a ln(bx) 42 = a ln (40 ÷ 63) 42 = a –0·454255 … (calculator)a = –92 (to nearest whole number)
Eiffel tower can be modelled by
y = –92 ln(x/63)
![Page 25: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/25.jpg)
As can be seen in the graph, there is a simple cubic function which for –1 > x > 1, and working in radians, behaves almost the same as the sine wave. i.e.
sin(x) px3 + qx2 + rx + s where p, q, r, and s are constants.
• The roots of this cubic are ±6 and 0. Express the cubic in terms of its factors viz. k(x – a)(x – b)(x – c).
• We know sin(π/6) = 1/2. Use this to find k as a simple fraction with a unit numerator.
-1.5
-1
-0.5
0
0.5
1
1.5
-3 -2 -1 0 1 2 3
sine
cubic
‘It fits where it touches.’
revealreveal
![Page 26: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/26.jpg)
As can be seen in the graph, there is a simple cubic function which for –1 > x > 1, and working in radians, behaves almost the same as the sine wave. i.e.
sin(x) px3 + qx2 + rx + s where p, q, r, and s are constants.
• The roots of this cubic are ±6 and 0. Express the cubic in terms of its factors viz. k(x – a)(x – b)(x – c).
• We know sin(π/6) = 1/2. Use this to find k as a simple fraction with a unit numerator.
-1.5
-1
-0.5
0
0.5
1
1.5
-3 -2 -1 0 1 2 3
sine
cubic
‘It fits where it touches.’
€
y = k(x − 0) x − 6( ) x + 6( )
= kx x 2 − 6( )
(a)
(b)
€
12 = k
π
6
π 2
36− 6
⎛
⎝ ⎜
⎞
⎠ ⎟≈ −3k
⇒ k = − 16
![Page 27: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/27.jpg)
A function is defined by f: x (x2 – x – 2)
Find the largest possible domain for the function.
revealreveal
![Page 28: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/28.jpg)
A function is defined by f: x (x2 – x – 2)
Find the largest possible domain for the function.
√(x 2 – x –
-1
0
1
2
3
4
5
6
-3 -2 -1 0 1 2 3 4 5
The function within the radical sign must be greater than or equal to zero.
€
x 2 − x − 2 ≥ 0
⇒ x − 2( ) x +1( ) ≥ 0
y=x2 – x – 2
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3 4 5
The sketch of this quadratic tells us that x ≥ 2 or x ≤ –1.
The sketch on the left shows the function in question.
![Page 29: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/29.jpg)
f(x) = 2x – 1 and g(x) = x2 + 2.
• Find an expression for f(g(x).
• In general f(g(x)) ≠g(f(x)).
However, in this case there are two values of x for which f(g(x)) = g(f(x)).
Find these values.
revealreveal
![Page 30: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/30.jpg)
f(x) = 2x – 1 and g(x) = x2 + 2.
• Find an expression for f(g(x).
• In general f(g(x)) ≠g(f(x)),
however, in this case there are two values of x for which f(g(x)) = g(f(x)).
Find these values.
(a) f(g(x) = f(x2 + 2)
= 2(x2 + 2) – 1
= 2x2 + 3
• g(f(x)) = g(2x – 1)
= (2x – 1)2 + 2
= 4x2 – 4x + 3
g(f(x)) = f(g(x)
(a) 4x2 – 4x + 3 = 2x2 + 3
(b) 2x2 – 4x = 0
(c) x(x – 2) = 0
(d) x = 0 or x = 2
![Page 31: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/31.jpg)
y = f(x)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2 -1 0 1 2 3
revealreveal
The sketch shows part of the function y = f(x)
(a) Draw a sketch of (i) y = f(–x) (ii) y = f(1 – x)
(b) Make a sketch of y = f´(x)
(0·75, 1)
![Page 32: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/32.jpg)
y = f(x)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2 -1 0 1 2 3
The sketch shows part of the function y = f(x)
(a) Draw a sketch of (i) y = f(–x) (ii) y = f(1 – x)
(b) Make a sketch of y = f´(x)
(0·75, 1)
y = f(-x)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
y = f(x)
y = f(-x)
(a (i))
y=f(1-x)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
y=f(x)
y=f(-x)
y=f(1-x)
(a (ii))
y=f'(x)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5
y=f(x)
y=f'(x)
(b)
![Page 33: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/33.jpg)
Q1 Find the maximum value of the function defined by:
€
f : x →1
x 2 + 4x + 9
Q2 Prove that y = 3x3 + 3x2 + 5x + 1
is an increasing function.
Where completing the square is useful
revealreveal
![Page 34: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/34.jpg)
Q1 Find the maximum value of the function defined by:
€
f : x →1
x 2 + 4x + 9
Q2 Prove that y = 3x3 + 3x2 + 5x + 1
is an increasing function.
Where completing the square is useful
In Higher maths you don’t know how to differentiate this function.
Complete the square on the denominator:
x2 + 4x + 9 = (x + 2)2 – 22 + 9 = (x + 2)2 + 5.
The smallest this expression can be is when the bracket takes the value zero …
When x = –2, the expression is worth 5.
This is when the function will be at its biggest.
f(–2) = 1/5
Q1
Q2 To prove the function is always increasing , you have to prove that the derivative is always positive.
€
dy
dx= 9x 2 + 6x + 5
Complete the square:
€
9x 2 + 6x + 5 = 9 x 2 + 69 x[ ] + 5 = 9 x 2 + 2
3 x[ ] + 5
= 9 x + 13( )
2− 1
3( )2
[ ] + 5 = 9 x + 13( )
2− 1
9[ ] + 5
= 9 x + 13( )
2−1+ 5
= 9 x + 13( )
2+ 4
The minimum value of the derivative is 4.Thus its always positive and thusAlways increasing.
![Page 35: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/35.jpg)
Solve the equation 2sin x = 1 for 0 ≥ x ≥ 4π
Many repetitive situations can be modelled by the sine function.Using suitable units the distance of the star from the centre of the picture can be modelled by D = 2 sin x where x is measured in radians.
revealreveal
![Page 36: Functions and Graphs.ppt](https://reader036.vdocument.in/reader036/viewer/2022081514/563dbc01550346aa9ab06fb8/html5/thumbnails/36.jpg)
Solve the equation 2sin x = 1 for 0 ≥ x ≥ 4π
Many repetitive situations can be modelled by the sine function.Using suitable units the distance of the star from the centre of the picture can be modelled by D = 2 sin x where x is measured in radians.
€
2sin x =1
⇒ sin x = 12
⇒ x = π6 or π − π
6 = 6π6 − π
6 = 5π6
or any number of complete revolutions more (or less) than these two solutions …
€
x = π6 or 5π
6 or 2π + π6 or 2π + 5π
6
or 4π + π6 or 4π + 5π
6
x = π6 , 5π
6 ,13π6 ,17π
6 , 25π6 , 29π
6
All other ‘answers’ are outside the desired range.