Core 3

FACTORISE AND CANCEL FACTORS

3x+6                 3(x+2)                3 

--------       =    -------------   =   --------------

x^2 -4              (x+2)(x-2)           x-2

ADD AND SUBTRACT BY FINDING A COMMON DENOMINATOR

3           2               3          2            6           2             4            2

-----  -  -----    =   ------- -  -------  =  ------  -  -------  =  -------- =  ---------

x+3     2x+6       x+3       2(x+3)    2(x+3)   2(x+3)    2(x+3)     x+3

Degree = highest power of a polynomial

Divisor = the thing you are dividing by

Quotient = the bit that you get when you divide by the divisor

(x^4 +x^2 -10) / (x^2 -2)        

                                x^2          +3

x^2 -2  l  x^4 +0x^3 +x^2 +0x -10

              x^4            -2x^2                               =x^2 + 3 +     4    

                                +3x^2                                               x^2 - 2

                                +3x^2      -6

                                                 4

1  +  1/y  x     y^2      (rationalise y multiplying by one)

1  -  1/y          y^2

=    y^2 +y 

      y^2 - 1

=  y(y+1)                    (factorise)

  (y+1)(y-1)

=      y                        (cancel)

      y -1

A mapping is an operation that takes one number and tranforms it into another.

The set of numbers you start with is called the domain, and the set of numbers they become is called the range.

Some mappings take every number in the domain to only one number in the range, these are functions. Also known as one-to-one or many-to-one functions.

Some mappings that aren't functions can be turned into functions by restricting their domain.

To restrict the domain of quadratics we complete the square

y = x^2 - 2x - 3    >>>>>>  y = (x-1)^2 - 4

domain =xER range = yER y>-4  >restrict the domain>  domain = xER x>1 range = yER y>-4

It is a one-to-one function now because none of the curve is repeated so now the inverse is valid.

Composite functions are two or more functions that have been combined for example fg(x) = f(g(x))

f(x) = √x g(x) = 1/(x-1)    solve the equation fg(x) = 1/2

fg(x) = f(1/(x-1)) = √1/(x-1)  =  1/√(x-1)  = 1/2

√(x-1)  = 2 >>>> x-1 = 4 >>>> x=5

The domain of fg(x) is x>1 because the denominator can't equal 0

Therefore the range is fg(x)>0   (sub in x=1)

Only one-to-one functions have inverse functions. Self inverse functions: f^-1(f(x)) = x

y =   x + 1 

        x - 2

x  =    y + 1       swap x and y

          y - 2

x(y-2) = y+1 >>> xy-2x = y+1 >>> xy-y = 2x+1 >>> y(x-1) = 2x+1

y =   2x + 1        solve to get y=

          x - 1

f^-1 =   2x +1      rewrite as f^-1

              x -1

The modulus of a number is it's size it doesnt matter if it is positive or negative.

|f(x)| = f(x) when f(x)>0 and -f(x) when f(x) <0 any negative values of f(x) are made positive by reflecting them in the x axis. This restricts the range of the modulus function to |f(x)| > 0.

For the graph of f(|x|) negative x values produce the same result as the corresponding positive x value. So the graph of f(x) for x > 0 is reflected in the y axis for negative values of x.

Solve |x^2 - 2x -3| = 1 - x

Sketch the graphs of the 2 functions which shows they cross twice.

Work out the ranges of x for which f(x)>0 and f(x)<0 by looking at where it crosses the axis

x^2 - 2x -3 = 1 - x  for x<-1 or x>3         x= -1.562

-(x^2 -2x -3) = 1-x  for -1<x<3               x= -0.562

Check against graph that they are what you expected and are valid.

y = f(x +c)  for c>0 f(x) is shifted c to the left, for c<0 f(x) is shifted c to the right.

y = f(x) +c  for c>0 f(x) is shifted c upwards, for c<0 f(x) is shifted c downwards.

y = af(x)   for a>1 f(x) is stretched vertically by a factor of a, if 0<a<1 f(x) is squashed (stretch factor 1/a), if a<0 f(x) is also reflected in the x axis.

y = f(ax)   for a>1 f(x) is squashed horizontally by a factor of a, if 0<a<1 the graph is stretched horizontally, if a<0 f(x) is also reflected in the y axis.

If you are combining transformations do the bit in the brackets first.

Arcsine is the inverse of sine, arccos is the inverse of cos and arctan is the inverse of tan.

The inverse trig functions reverse the effect of trig, e.g. sin30 = 0.5 and arcsin0.5 = 30

The trig functions are not one-to-one mapping so to find the inverse you have to restrict the domain. The inverse graphs are reflections of the original graphs in the line y=x.

Cosec is the reciprocal of sin (1/sin), sec is the reciprocal of cos (1/cos) and cot is the reciprocal of tan (1/tan), since tan = sin/cos cot = cos/sin.

Cosecx is undefined at any point where sinx=0, secx is undefined at any point where cosx=0 and cotx is undefined at any point where tanx=0.

The graphs have minimum points whenever their reciprocal has a maxium.

cotx crosses the x axis whenever tanx has an asymptote - (npi +pi/2, 0)

Secx has asymptotes when cosx = 0 - (npi + pi/2) = x

Cosecx has asymptotes whenever sinx=0 - x = npi

cos^2 x + sin^2 x = 1

sec^2 x = 1 + tan^2 x

cosec^2 x = 1 + cot^2 x

sin(A +- B) = sinAcosB +- cosAsinB

cos(A +- B) = cosAcosB -+ sinAsinB

tan(A +- B) =   tanA +- tanB                                                                                                                2222222222211 -+ tanAtanB

sin2A = 2sinAcosA

cos2A = cos^2A-sin^2A = 2cos^2A - 1 = 1 - 2sin^2A

tan2A =       2tanA           aaaa1 - aaaaaaaaa1-aaaaaaaa1 aaaaaa1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-tan^2A                                                                                                       

Express 2sinx - 3cosx in the form Rsin(x-a)  

Rsin(x-a) = R(sinxsina - cosxsina)      Expand using the addition formula

Rcosa = 2         Rsina =3                    Equate the coefficients of sinx and cosx

Rsina/Rcosa = tana = 3/2                   Divide sin by cos to get tan

a= tan^-1 3/2 = 56.31 degrees            Take tan^-1 of the result to find a

R^2 = 2^2 + 3^2 = 13 >>> R=√13       To find R square the equations and add them together then

so 2sinx - 3cosx = √13 sin (x-56.31)    square root and write the full answer out

The maximum value is +R the minimum value is -R

The gradient of y=e^x is e^x. y=e^x cuts the y axis at (0,1). e^x can't be zero or negative.

lnx is the inverse function of e^x, it is a reflection of e^x is the line y=x, it cuts the x axis at (1,0)

Log Rules

lnx + lny = ln(xy)

lnx - lny = ln(x/y)

lnx^k = klnx

N= Noe^(at)  is an equation that models the growth of organisms in a river, N=Number of organisms per unit volume, t=time in weeks since start of observation, No and a are constants.      

After 4 weeks N is double, find a? N=2No >>> 2No=Noe^4a >>> 2=e^4a >>> ln2=4alne>>>ln2=4a >>>a= 0.1733

No =20, what is N after 10 weeks? N= 20e^(0.1733x10)>>> N=113.1

How many weeks for N to treble the initial value? N=3No>>> 3No=Noe^(0.1733t)>>>3=e^(0.1733t)>>>ln3=0.1733t>>>ln3/0.1733=t=6.339>>> to the nearest week = 6 weeks

Give a reason why this mode of growth is unrealistic

It can't keep growing at that rate because the river would fill up.

if y = f(g(x)) >>> dy/dx = f'(g(x)) g'(x)

y = (2x+7)^10  dy/dx = 10(x+7)^9 x 2 = 20(x+7)^9

y = cosx^2  dy/dx = -sin(x^2) x 2x = -2xsin(x^2)

y= 1/ √(x^2+4x) = (x^2+4x)^-(1/2)   dy/dx = -1/2 (x^2+4x)^(-3/2) x (2x+4) = -(x+2)/√(x^2+4x)

y = e^x    dy/dx = e^x

y = e^x^2 + 2e^x     dy/dx = 2xe^x^2 + 2e^x

y = lnx    dy/dx = 1/x

y = ln(x^2 + 3)  dy/dx = 2x/(x^2 + 3)

y = ln(f(x))  dy/dx= f'(x)/f(x)

sinx >>>> cosx

cosx >>>> -sinx

tanx >>>> sec^2 x

Proove using quotient rule:

cosecx >>>> -cosecxcotx

secx >>>> secxtanx

cotx >>>> -cosec^2 x

If y = f(x)g(x)   dy/dx = f(x)g'(x) + f'(x)g(x)

y = (x^2 +7)(x^4 - 5)

f(x) = x^2 +7    f'(x) = 2x

g(x) = x^4 - 5   g'(x) = 4x^3

dy/dx = (x^2 +7)4x^3  +  2x(x^4 - 5)

         =  4x^5 + 28x^3 +2x^5 - 10x

         =  6x^5 + 28x^3 - 10x

y = f(x)/g(x)    dy/dx = (gf' - fg')/g^2

y = tanx = sinx/cosx

f(x) = sinx   f'(x) = cosx

g(x) = cosx     g'(x) = -sinx

dy/dx = (cosxcosx - sinx (-sinx))/cos^2 x >>> (cos^2 x + sin^2 x) /cos^2 x = 1/cos^2 x

= sec^2 x

Equation of the tangent:

Differentiate and sub in coordinates to find the gradient (m). Use this value to find the equation of the line by putting m and the coordinates into the equation y-y1=m(x-x1).

Turning Points and Stationary Points

Stationary points occur when dy/dx = 0

To determine whether it is a minimum or a maximum find the second derivative d2y/dx2

d2y/dx2<0 is a maximum point

d2y/dx2>0 is a minimum point

Show that f(x)= x^4 + 3x - 5 has a root in the interval 1.1<x<1.2.

f(1.1) = -0.2359     f(1.2) = 0.6736

Since f(x) is a continuous variable and f(1.1)<0 and f(1.2)>0 the root is in the interval [1.1,1.2]

Show that the equation f(x) = 25x^2e^(2x) - 16 = 0 can be written as +- 4/5 e^-x

2xx^2e^(2x)-16=0>>>25x^2e^(2x)=16>>>25x^2=16/(e^(2x))>>>x^2=16/(25e^(2x))>>>                x=√{16/(25e^(2x)} = ±4/(5e^x) = ±4/5 e^-x

This is can be used as an iteration formula to find roots

Xn+1 = 4/5 e^(-Xn)   Xo = 0.5 calculate X1, X2 and X3 to 3dp

X1 = 4/5 e^(-Xo) = 0.485

X2 = 0.492

X3 = 0.489

Give an accurate esitimate to 2dp (They all round to 0.49 to 2dp)f(0.485)= -0.487f(0.495) = 0.485There is a sign change between f(upper bound) and f(lower bound) and f(x) is a continuous variable so the root must be 0.49 to two decimal places.