3.17 Composition of functions (exercise)  (Page 2/2)

$$f\left(\left|x\right|\right)=\left|x\right|-1;-2\le x\le 2$$

Problem 4: Two functions are given as :

$$f\left(x\right)=-1+|x-1|;-1\le x\le 3$$

$$g\left(x\right)=2-|x+1|;-2\le x\le 2$$

Find fog(x).

Solution :

Statement of the problem : The domains of the given functions are different. We need to determine composition and interpret domain of the composition.

$$\text{Let}y=g\left(x\right)=\left(2-|x+1|\right);-2\le x\le 2$$

Now, let us first determine the composition,

$$\Rightarrow fog\left(x\right)=f\left(y\right)=f\left(2-|x+1|\right)|=-1+|2-|x+1|-1|$$

This is the rule of the composition function. In order to find the domain of the composition, we write domains of two functions as an intersection :

$$\Rightarrow fog\left(x\right)=-1+|2-|x+1|-1|;-1\le y\le 3\text{and}-2\le x\le 2$$

We interpret the interval “-1 ≤ y ≤ 3 “ as :

$$-1\le \left(2-|x+1|\right)\le 3\Rightarrow -3\le \left(-|x+1|\right)\le 1$$

Multiplying each term with “-1” and reversing inequality, we have :

$$3\ge \left(|x+1|\right)\ge -1$$

But, "|x+1| ≥ -1" is true for all values of “x”. Hence, above inequality is equal to interval given by first part "|x+1| ≤ 3" as :

$$\Rightarrow -3\le x+1\le 3$$

$$\Rightarrow -4\le x\le 2$$

Hence, interval of the composition is intersection of intervals “-4≤ x ≤2“ and “-2 ≤ x ≤ 2”.

$$\text{Domain}=-2\le x\le 2$$

The composition, therefore, is :

$$\Rightarrow fog\left(x\right)=-1+|1-|x+1\left|\right|;-2\le x\le 2$$

Problem 5: Two functions are defined as :

$$g\left(x\right)=\frac{1-x}{1+x};0\le x\le 1$$

$$h\left(x\right)=4x\left(1-x\right);0\le x\le 1$$

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

$$y=h\left(x\right)=4x\left(1-x\right)=4x-4x^2;0<=x<=1$$

Then,

$$\Rightarrow goh\left(x\right)=g\left(y\right);0\le y\le 1\text{and}0\le x\le 1$$

$$\Rightarrow goh\left(x\right)=g(4x-4x^2);0\le 4x-4x^2\le 1\text{and}0\le x\le 1$$

Let us first interpret rule of the function :

$$\Rightarrow goh\left(x\right)=\frac{1-y}{1+y}=\frac{1-\left(4x-4x^2\right)}{1+\left(4x-4x^2\right)}=\frac{1-4x+4x^2}{1+4x-4x^2}$$

Now, we interpret the interval “ $0\le \left(4x-4x^2\right)\le 1$ ” in parts. The left part is :

$$\left(4x-4x^2\right)\ge 0\Rightarrow x\left(1-x\right)\ge 0\Rightarrow x\ge 0,x\le 1\Rightarrow 0\le x\le 1$$

The right part of the interval is :

$$\left(4x-4x^2\right)\le 1\Rightarrow \left(4x^2-4x\right)\ge -1\Rightarrow \left(4x^2-4x+1\right)\ge 0$$

Applying sign scheme, this inequality is valid for all values of “x” :

$$\Rightarrow -\infty \le x\le \infty$$

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 :

$$\Rightarrow goh\left(x\right)=\frac{1-4x+4x^2}{1+4x-4x^2};0\le x\le 1$$

For determining hog(x), Let

$$y=g\left(x\right)=\frac{1-x}{1+x};0\le x\le 1$$

Then,

$$\Rightarrow hog\left(x\right)=g\left(y\right);0\le y\le 1\text{and}0\le x\le 1$$

$$\Rightarrow hog\left(x\right)=g\left(\frac{1-x}{1+x}\right);0\le \left(\frac{1-x}{1+x}\right)\le 1\text{and}0\le x\le 1$$

Let us first interpret rule of the function :

$$\Rightarrow hog\left(x\right)=4y\left(1-y\right)=4\frac{1-x}{1+x}\left(1-\frac{1-x}{1+x}\right)$$

$$\Rightarrow hog\left(x\right)=4\frac{1-x}{1+x}\left(\frac{1+x-1+x}{1+x}\right)=\frac{8x\left(1-x\right)}{{\left(1+x\right)}^2}$$

Now, we interpret the interval “ $0\le \left(\frac{1-x}{1+x}\right)\le 1$ ” in parts. The left part is :

$$\frac{1-x}{1+x}\ge 0\Rightarrow 1-x\ge 0$$

Applying sign scheme, this inequality is valid for all values of “x” :

$$\Rightarrow 0\le x\le 1$$

The right part of the interval is :

$$\frac{1-x}{1+x}\le 1\Rightarrow 1-x\le 1+x\Rightarrow 0\le 2x\Rightarrow x\ge 0$$

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 :

$$\Rightarrow hog\left(x\right)=\frac{8x\left(1-x\right)}{{\left(1+x\right)}^2};0\le x\le 1$$

Problem 6: A function is defined as :

| 1 + x ; x≥0 f(x) = || 1- x ; x<0

Determine composition f{f(x)}.

Solution :

Statement of the problem : The function is defined by different rules in two intervals.

For f(1+x) in the interval x>=0

| 1 + 1+ x ; 1+x ≥0 and x≥0 f(1+x) = || 1- 1 - x ; 1+x<0 and x≥0

| 2+ x ; x ≥-1 and x≥0 f(1+x) = || - x ; x<-1 and x≥0

The intersection of upper intervals “x ≥-1 and x≥0” is equal to “x≥0”. There is no common interval for the intersection of lower intervals. Hence,

$$f\left(1+x\right)=2+x;x\ge 0$$

For f(1-x) in the interval x<0

| 1 + 1- x ; 1-x ≥0 and x≤0 f(1-x) = || 1- 1 + x ; 1- x<0 and x<0

| 2- x ; x ≤1 and x<0 f(1-x) = || x ; x>1 and x<0

The intersection of upper intervals “x ≤1 and x<0” is equal to “x<0”. There is no common interval for the intersection of lower intervals. Hence,

$$f\left(1-x\right)=2-x;x<0$$

Therefore, the composition is :

| 2+x ; x>0 f{f(x)}= || 2- x ; x<0