Problem 5: Two functions are defined as :
g
x
=
1
−
x
1
+
x
;
0
≤
x
≤
1
g
x
=
1
−
x
1
+
x
;
0
≤
x
≤
1
h
x
=
4
x
1
−
x
;
0
≤
x
≤
1
h
x
=
4
x
1
−
x
;
0
≤
x
≤
1
Determine the composition goh(x) and hog(x).
Solution :
Statement of the problem : The domains of the given functions are same. We need to interpret domain of the composition as we compose the required function.
Let
y
=
h
x
=
4
x
1
−
x
=
4
x
−
4
x
2
;
0
<
=
x
<
=
1
y
=
h
x
=
4
x
1
−
x
=
4
x
−
4
x
2
;
0
<
=
x
<
=
1
Then,
⇒
g
o
h
x
=
g
y
;
0
≤
y
≤
1
and
0
≤
x
≤
1
⇒
g
o
h
x
=
g
y
;
0
≤
y
≤
1
and
0
≤
x
≤
1
⇒
g
o
h
x
=
g
(
4
x
−
4
x
2
)
;
0
≤
4
x
−
4
x
2
≤
1
and
0
≤
x
≤
1
⇒
g
o
h
x
=
g
(
4
x
−
4
x
2
)
;
0
≤
4
x
−
4
x
2
≤
1
and
0
≤
x
≤
1
Let us first interpret rule of the function :
⇒
g
o
h
x
=
1
−
y
1
+
y
=
1
−
4
x
−
4
x
2
1
+
4
x
−
4
x
2
=
1
−
4
x
+
4
x
2
1
+
4
x
−
4
x
2
⇒
g
o
h
x
=
1
−
y
1
+
y
=
1
−
4
x
−
4
x
2
1
+
4
x
−
4
x
2
=
1
−
4
x
+
4
x
2
1
+
4
x
−
4
x
2
Now, we interpret the interval “
0
≤
4
x
−
4
x
2
≤
1
0
≤
4
x
−
4
x
2
≤
1
” in parts. The left part is :
4
x
−
4
x
2
≥
0
⇒
x
1
−
x
≥
0
⇒
x
≥
0
,
x
≤
1
⇒
0
≤
x
≤
1
4
x
−
4
x
2
≥
0
⇒
x
1
−
x
≥
0
⇒
x
≥
0
,
x
≤
1
⇒
0
≤
x
≤
1
The right part of the interval is :
4
x
−
4
x
2
≤
1
⇒
4
x
2
−
4
x
≥

1
⇒
4
x
2
−
4
x
+
1
≥
0
4
x
−
4
x
2
≤
1
⇒
4
x
2
−
4
x
≥

1
⇒
4
x
2
−
4
x
+
1
≥
0
Applying sign scheme, this inequality is valid for all values of “x” :
⇒

∞
≤
x
≤
∞
⇒

∞
≤
x
≤
∞
The intersection of two parts is “0≤x≤1”. Thus, interval of composition is intersection of intervals “0≤x≤1” and “0≤x≤1”, which is “0≤x≤1”. Therefore, “goh(x)” is :
⇒
g
o
h
x
=
1
−
4
x
+
4
x
2
1
+
4
x
−
4
x
2
;
0
≤
x
≤
1
⇒
g
o
h
x
=
1
−
4
x
+
4
x
2
1
+
4
x
−
4
x
2
;
0
≤
x
≤
1
For determining hog(x), Let
y
=
g
x
=
1
−
x
1
+
x
;
0
≤
x
≤
1
y
=
g
x
=
1
−
x
1
+
x
;
0
≤
x
≤
1
Then,
⇒
h
o
g
x
=
g
y
;
0
≤
y
≤
1
and
0
≤
x
≤
1
⇒
h
o
g
x
=
g
y
;
0
≤
y
≤
1
and
0
≤
x
≤
1
⇒
h
o
g
x
=
g
1
−
x
1
+
x
;
0
≤
1
−
x
1
+
x
≤
1
and
0
≤
x
≤
1
⇒
h
o
g
x
=
g
1
−
x
1
+
x
;
0
≤
1
−
x
1
+
x
≤
1
and
0
≤
x
≤
1
Let us first interpret rule of the function :
⇒
h
o
g
x
=
4
y
1
−
y
=
4
1

x
1
+
x
1

1

x
1
+
x
⇒
h
o
g
x
=
4
y
1
−
y
=
4
1

x
1
+
x
1

1

x
1
+
x
⇒
h
o
g
x
=
4
1

x
1
+
x
1
+
x

1
+
x
1
+
x
=
8
x
1
−
x
1
+
x
2
⇒
h
o
g
x
=
4
1

x
1
+
x
1
+
x

1
+
x
1
+
x
=
8
x
1
−
x
1
+
x
2
Now, we interpret the interval “
0
≤
1
−
x
1
+
x
≤
1
0
≤
1
−
x
1
+
x
≤
1
” in parts. The left part is :
1

x
1
+
x
≥
0
⇒
1
−
x
≥
0
1

x
1
+
x
≥
0
⇒
1
−
x
≥
0
Applying sign scheme, this inequality is valid for all values of “x” :
⇒
0
≤
x
≤
1
⇒
0
≤
x
≤
1
The right part of the interval is :
1
−
x
1
+
x
≤
1
⇒
1
−
x
≤
1
+
x
⇒
0
≤
2
x
⇒
x
≥
0
1
−
x
1
+
x
≤
1
⇒
1
−
x
≤
1
+
x
⇒
0
≤
2
x
⇒
x
≥
0
The intersection of two parts is “0≤x≤1”. Thus, interval of composition is intersection of intervals “0≤x≤1” and “0≤x≤1”, which is “0≤x≤1”. Therefore, “goh(x)” is :
⇒
h
o
g
x
=
8
x
1
−
x
1
+
x
2
;
0
≤
x
≤
1
⇒
h
o
g
x
=
8
x
1
−
x
1
+
x
2
;
0
≤
x
≤
1