Pure Core 3
A key point Summary of what's involved in each part of core three
- Created by: Ellen Shercliff
- Created on: 25-05-11 17:55
Functions 1/2
- A function is a one-one or many-one mapping
- The set of numbers for which a function is defined is called the domain
- A function of f consists of:defining a rule, its domain
- The set of values a function takes is called the range
- When the domain of f is a continous interval, the range can be found by considering the graph of y=f(x). The range consists of the values y can take and is written as an inequality using f(x)
Functions 2/2
- the composite function fg means f(g(x))
- a function only has an inverse where it is one to one, and is obtained by reflecting f(x) in y=x to get f^-1(x)
- A reverse flow diagram can be used to obtain the inverse of a function when x occurs only once in f(x).
- The inverse of f can be found my writing y=f(x), rearranging the equations so you get x= something and substituting x and y
Tranforming graphs 1/2
- a translation of [a]
[b] translates the graph of y= f(x) to y=f(x-a) +b
- The graph of y=f(x) is transformed into the graph of y=-f(x) by a reflection in the x axis
- The graph of y=f(x) is transformed into the graph of y=f(-x) by reflection in the y axis
Transforming graphs 2/2
- The graph of y=f(X) is transformed onto the graph of y=df(x) by a stretch scale factor d in the y-direction
- the graph of y=f(x) is transformed onto y=f(x/c) by a stretch of scale factor c in the x-direction
- The modulus function |x| is defined by:
>|x|=x when x>/= 0
>|x|=-x when x<0
Inverse Trigonometric equations 1/2
- Sin^-1x has a domain -1 to 1 and a range -pi/2 to pi/2
- Cos^-1x has a domain -1 to 1 and a range 0 to pi
- Tan^-1x has a domain all real number and a range -pi/2 to pi/2
Inverse Trigonometric equations 2/2
- secx=1/cosx
- cosecx=1/sinx
- cotx= 1/tanx or cosx/sinx
- cos^2X + sin^2X=1
- 1+tan^2X=sec^2X
- cot^2X+1=cosec^2X
e 1/2
- The number e is irrational and its value is approximately 2.718
- The curve with the equation y=e^x has derivative dy/dx= e^x
- d/dx e^kx=ke^kx
- ∫ekx= 1/ke^kx (+c)
e 2/2
- The natural logarithm ln x is definied for x>0 A useful equivelent statement is:
x=e^x <==> y=lnx - The derivative of ln x is 1/x
- ∫1/xdx= lnx +c
Further differentiation and the chain rule 1/2
- y=sinx dy/dx= cos x
- y=cosx dy/dx=-sinx
- the chain rule can be used to differentiate composite functions:
dy/dx= dy/du X du/dx
Further differentiation and the chain rule 2/2
- The chain rule can be extended to:
dy/dx= dy/dv X dv/du X du/dx
- dy/dx= 1/dx/dy and dx/dy=1/dy/dx
Differentiation using the product and quotient rul
- where y=uv,
dy/dx=udv/dx+vdu/dx - where y=u/v,
dy/dx=(vdu/dx-udv/dx)/v^2
Differentiation using the product and quotient rul
- y=tanx dy/dx= sec^2X
- y=cotx dy/dx= -cosecx
- y=secx dy/dx= secxtanx
- y=cosecx dy/dx= -cosecxcotx
Numerical solution of equations and iterative meth
- If the graph of y=f(x) is continuous over the interval a to b and f(a) and f(b) have different signs, then at least one root of the equations f(x)=0 must lie in the interval a to b
- Cobweb and/or staircase diagrams can be drawn to illustrate whether convergence takes place or not for iterations of form x(n+1)=g(xn)
Numerical solution of equations and iterative meth
An iteration of the form x(n+1)=g(xn) converges when the gradient of y=g(x) at the point of the intersection with the line y=x satisfies the condition |g'(x)|<1, provided a suitable value for x1 is chosen.
integration by inspection and substitution 1/2
- for constants a,b,n where n =/=-1
∫(ax+b)^(n)dx=1/a(n+1) (ax+b)^(n+1) (+c)
- ∫e^(ax+b)dx= 1/ae^(ax+b) (+c)
- for x in radians,
∫cosxdx= sinx (+c)
∫sinxdx= -cosx (+c)
∫sec^2xdx= tanx (+c)
- for constants a,b and x
∫cos(ax+b)dx= 1/a sinx(ax+b) (+c)
∫sin(ax+b)dx= -1/a cos(ax+b) (+c)
∫sec^2(ax+b)=1/a tan(ax+b) (+c)
integration by inspection and substitution 2/2
- for constants a and x
∫cosecax cotax dx =-1/a cosecax (+c)
∫secax tanax dx=1/a secax (+c)
∫cosec^2ax dx=-1/a cotax (+c)
∫tanax dx= 1/a ln |secax| (+c)
∫cotax dx= 1/a ln|sinax| (+c)
∫secax dx= 1/a ln|secax+tanax| (+c)
∫cosecax dx=-1/a ln|cosecax + cotax| (+c)
- ∫f'(x)/f(x)dx= ln|f(x)| (+c)
Integration by parts and standard integrals 1/2
- ∫u dv/dx dx= uv- ∫v du/dx dx is known as the integration by parts formula and is given in the formula book.
- It is used to solve:
∫x^n sinmx dx
∫x^n cosmx dx
∫x^n e^mx dx
∫x^n ln mx dx
Integration by parts and standard integrals 2/2
∫1/(a^2+x^2)dx= 1/atan^-1(x/a) (+c)
∫1/√(a^2+x^2)dx=
sin^-1 (x/a) (+c)
Volume of revolution and numerical integration 1/2
- When a region bound by the curve y=f(X), the x axis, the lines x=a and x=b is rotated through 2pi radians about the x-axis the volume, v of the solid generated is called the volume of revolution and is given by:
v=∫(between a and b) piy^2dx= ∫(between a and b)pi[f(x)]^2dx
- When a region bound by the curve x=g(y), the y axis, the lines y=c and y=d is rotated through 2pi radians about the y-axis the volume, v of the solid generated is called the volume of revolution and is given by:
v=∫(between c and d) pix^2dx= ∫(between c and d)pi[g(y)]^2dx
Volume of revolution and numerical integration 2/2
- In all the numerical methods of integration an improvement on the estimate can be obtained by increasing the number if strips
- The mid-ordinate rule for n strips is:
∫(between a and b) ydx =h(y1/2+y3/2+...+y(n-3/2)+y(n-1/2))
where h=b-a/n
- Simpsons rule for n strips and (n+1) ordinates where n is even is:
∫(between a and b) ydx =1/3h[(y0+yn)+4(y1+y3+...+y(n-1))+2(y2+y4+...+y(n-2)]
where h=b-a/n
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