11 x1 t12 07 primitive function (2013)

The Primitive FunctionIf we know the equation of the tangent, how do we find the original
curve?

The Primitive Function
 
  c
n
x
xf
xxf
n
n





1
is;functionprimitivethen the,If
1
If we know the equation of the tangent, how do we find the original
curve?

 
  c
n
x
xf
xxf
n
n





1
1
    4
3e.g. xxfi 
curve?

 
  c
n
x
xf
xxf
n
n





1
1
    4
3e.g. xxfi 
curve?
  c
x
xf 
5
3 5

 
  c
n
x
xf
xxf
n
n





1
1
    4
3e.g. xxfi 
curve?
  c
x
xf 
5
3 5
    256 23
 xxxxfii

 
  c
n
x
xf
xxf
n
n





1
1
    4
3e.g. xxfi 
curve?
  c
x
xf 
5
3 5
    256 23
  cx
xxx
xf  2
23
5
4
6 234

 
  c
n
x
xf
xxf
n
n





1
1
    4
3e.g. xxfi 
curve?
  c
x
xf 
5
3 5
    256 23
  cx
xxx
xf  2
23
5
4
6 234
  cxxxxxf  2
2
1
3
5
2
3 234

(iii) Find the equation of the curve which passes through (1,1) and has
a gradient function of 2x + 3

32  x
dx
dy

32  x
dx
dy
cxxy  32

32  x
dx
dy
cxxy  32
when x = 1, y = 1

32  x
dx
dy
cxxy  32
3
311i.e. 2


c
c
when x = 1, y = 1

32  x
dx
dy
cxxy  32
3
311i.e. 2


c
c
332
 xxy
when x = 1, y = 1

32  x
dx
dy
cxxy  32
3
311i.e. 2


c
c
332
 xxy
when x = 1, y = 1
Exercise 10J; 1ace etc, 2bdf, 3aceg, 4bd, 5b, 7ac, 8bdf
9ace, 10b, 12bd, 14a, 15, 17a

11 x1 t12 07 primitive function (2013)

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