function theory

BY: M. PRAHAS TOMI M. S.

XI SC-7SMAN 1 BOGOR

2009

M. PRAHASTOMI M. S.

BASIC EXPLANATION SOME CASES ‘THIS FOR YOU’ SECTION

M. PRAHASTOMI M. S.

FUNCTION COMPOSITION INVERS OF FUNCTION INVERS ALGORITHM INVERS OF FUNCTION COMPOSITION

M. PRAHASTOMI M. S.

It Means the function of certain function(s) which given.

Simply we write the function composition above:

= (f o g o h o …) (x) = f [g {h (… (x)}]

M. PRAHASTOMI M. S.

f(x) = ax + bg(x) = dx + ch(x) = mx + n

Case 1 (f o g) (x) = f [g(x)] = a [g(x)] + b = a (dx + c) + b

Case 2 (g o f) (x) = g [f(x)] = d [f(x)] + c = d (ax + b) + c

Case 3( f o g o h) (x) = f[ g{h(x)}] = f[g{mx + n}] = f[ d(mx + n) + c ] = a {d(mx + n) +c} + b

M. PRAHASTOMI M. S.

Given: f(ax + b) = mx + ng(qx +r) = wx + z

(f o g) (x) = f[g(x)]

Firstly, we’ve ta find f (x). f(x) ≠ f(ax + b)

so, f [{a.(x-b)/a }+ b] = {m (x-b)/a} + n why (x-b)/a replacing x which suppose to be there?

Ex, y = ax + b

x= (y-b)/a

We change y become x, and replace it into the equation:f(x) = m { (x-b)/a } + n

M. PRAHASTOMI M. S.

The same way with finding g (x)

g(x) = w{(x-r)/q} + n

so, (f o g) (x) = f[ g(x) ] = m [ {g(x) –b}/a ] + n

Algebraic count gives= f [g(x)] = (m/a)[{w(x-r) + q (b +z)}/q] + n

M. PRAHASTOMI M. S.

f(ax+b)= [cx +d]c+d

g(wx+y)= [px + q]p+q

h(mx+n)= [rx + s]r+s

(g o h o f) (x) =…

M. PRAHASTOMI M. S.

f(x) = ax + b

(f o g) (x) = cx + d

Find g(x) !

f[ g(x) ] = cx + d

f[ g(x) ] = cx+ d = a{g(x)} + b

g(x) = (cx + d –b)/a

M. PRAHASTOMI M. S.

f(x) = ax + b

(g o f) (x) = cx + d

g(x) =...

g[ f(x) ] = cx +d

g[ ax + b ] = cx + d

We can find g(x) with 4th case way

g(x) = c [(x-b)/a] + d

M. PRAHASTOMI M. S.

Symbolized: f -1 (x)

(f o f -1 ) (x) = ( f -1 o f ) (x)= x

M. PRAHASTOMI M. S.

f(x) = ax + b

f -1 (x) = …

(f o f -1 ) (x) = f [f -1 (x) ] = a[f -1 (x) ] + b = x

f -1 (x) = (x-b)/a

M. PRAHASTOMI M. S.

f(ax + b) = cx + d

f -1 (x) = …

f(x) = c{(x-b)/a} + d

(f o f -1 ) (x) = f [f -1 (x) ] = c{( f -1 (x) -b)/a} + d =x

f -1 (x) = a{(x-d)/c} + b

M. PRAHASTOMI M. S.

f(x) = [ (ax + b)/(cx + d) ]t

f -1 (x) = …

(f o f -1 ) (x) = f [f -1 (x) ] = [ {a f -1 (x) + b}/{c f -1 (x) + d} ]t = x

x 1/t = {a f -1 (x) + b}/{c f -1 (x) + d}

cx1/t f -1 (x) + d x 1/t = a f -1 (x) + b

f -1 (x) = (-d x 1/t + b)/ (c x 1/t –a)

M. PRAHASTOMI M. S.

f(x) = xLog (ax)/d f -1 (x) = …

(f o f -1 ) (x) = f [f -1 (x) ] = xLog [af -1 (x) ] / d = x

x x = [af -1 (x) ] / d

f -1 (x) = (dxx)/a

M. PRAHASTOMI M. S.

f(x) = m Log [(ax-b)/( b -mn – c h )]ax

f -1 (mx) = …

M. PRAHASTOMI M. S.

This is how computer know and count any function(s).Given case:f(x) = kx + p function algorithm 1 2x → kx → kx + pinverse algoritm 2 1x → x – p → (x-p)/k

Find the pattern of the following number!

M. PRAHASTOMI M. S.

f(x) = (dxn + wz)pq

inverse algoritm?

x → x 1/pq → x1/pq – wz → (x 1/pq – wz)d → [(x1/pq – wz)d] 1/n

M. PRAHASTOMI M. S.

f(x) = [rt(dxn + wz)pq] m

inverse algoritm?

x → x 1/m → (x 1/m )/rt → {(x 1/m )/rt}1/pq → {(x 1/m )/rt}1/pq – wz → [{(x 1/m )/rt}1/pq – wz]/d → [[{(x 1/m )/rt}1/pq – wz]/d] 1/n

M. PRAHASTOMI M. S.

Figure out inverse of logarithm!

f -1 (2x-c) = 3p[rt +k(dxhn + wz)pq] m-p+2t

M. PRAHASTOMI M. S.

for any given equation:

(f o g o h o …) -1 (x) = (... o h -1 o g -1 o f -1 ) (x)

M. PRAHASTOMI M. S.

f(x) = ax + d

g(x) = mxn + p2/n

(f o g) -1 (x) = ...

g -1 (x) = [(x – p2/n)m] 1/n

f -1 (x) = (x –d)/a

(f o g) -1 (x) = (g -1 o f -1 ) (x) = g -1 [f -1 (x)]

= [(f -1 (x) – p2/n)m] 1/n

= [{(x –d)/a } - p2/n }m] 1/n

M. PRAHASTOMI M. S.

f -1 (cx+d) = n(abx + mh+t ) m/n

g(x) = 3rx Log (ax)/d

Figure out (f -1 o g) -1 (x) !

M. PRAHASTOMI M. S.

See you!

M. PRAHASTOMI M. S.