Plotting the graph of f o g, and f o f
by using the graph of f and g.
By: Ahmad Ghandehari
1- The graph of function g is shown in figure (1) we want to find a point on graph of g
such that its abscissa is g(x1). x belongs to domain of g. For this purpose we draw the
line y  x in the figure (1).
A  g(x 1 ), g(x 1 )
Figure (1)
We select a point with abscissa the x1. on the x – axis. By drawing a connector parallel
to the y – axis, we find point A on the curve of g, so A(x1, g(x1)).
From point A we draw a connector parallel to the x – axis such that this connector
intersects the line y  x at point B and intersects the y – axis at point H. Then we can
write.
x B  yB  g(x1 )
And
yH  g(x1 )
From point B, we draw a connector, parallel to the y – axis such that this connector,
intersects the curve of g at point A.
The abscissa of A is g(x1), therefore
A  g(x 1 ), g(g(x 1 ))  .
2- The graphs of two functions f and g are shown in figure (2). We want to find a point
on curve of f such that its abscissa is g(x1).
x1  D g
For this purpose, we draw the line y  x on the figure (2).
Suppose the point A(x1, g(x1 )) is on the curve of g, we draw a connector parallel to the
x–axis until intersects the line y  x at point B, and intersects the y – axis at point H,
and then we can write
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x B  yB  g(x1 )
and
yH  g(x1 )
A  g(x 1 ), f(g(x 1 )) 
Figure (2)
From point B we draw a connector parallel to the y – axis until intersects the curve of f
at point A so A  g(x 1 ), f(g(x 1 ))  .
3- The graph of two functions f and g are shown in figure (3), we want to find a point
like A such that its coordinates are x1 , f(g(x 1 ))  .
For this intention we draw the line y=x on the figure (3), we consider that the point
Ax1, g(x 1 ) is on the curve of g.
Ax1, g(x 1 ) .
Ag(x 1 ), f(g(x 1 ))  .
Ax1, f(g(x 1 )) .
Figure (3)
With using article 2, we can find a point A such that: Ag(x 1 ), f(g(x 1 ))  .
Now from the point A on the curve of f, we draw a connector parallel to the x – axis
until intersects the connector which drew from point A parallel to the y – axis. This
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intersection point is A,so x A  x1 and yA  f (g(x1 )) , therefore Ax1, f (g(x1 ))  , so
point A is a point of composite function fog.
Notice:
In the figure (4), three points A, B and C are on the curve of function fog.
Ax1, f g(x1 )
Bx 2 , f g(x 2 )
Cx 3 , f g( x 3 ) 
Figure (4)
4- We want plotting the graph of fof, by using the graph of function f.
Suppose we have drawn the graph of function f in figure (5).
We draw the line y = x in the figure, then we find a point with abscissa x 1 on the curve
of f like, Ax1, f (x1 ) .
Ax1, f (x1 )
Af (x1 ), f (x1 )
Af (x1 ), f f (x1 )
Ax1, f f (x1 )
Figure (5)
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From point A, we draw a connector parallel to the x – axis until intersects the line y = x
at Af (x1 ), f (x1 ) and intersects the y – axis at point H, so yH  f (x1 ) .
From point A we draw a connector parallel to the y – axis until intersects the curve of f
at point A , therefore x A  f (x1 ) and yA  f f (x1 ) , so Af (x1 ), f f (x1 ) .
Now we draw a connector from point A parallel to the x – axis until intersects the y –
axis at point H which yH  f f (x1 ) , also this new connector intersects the connector
which have drawn before to find point A, at point A, so x A  x1 and yA  f f (x1 )
or Ax1, f f (x1 ).
In the figure (6), we have found three points A , B and C which they are on the
curve of function fof.
Ax1, f f (x1 )
Bx 2 , f f (x 2 )
Cx 3 , f f ( x 3 )
Figure (6)
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