5.9 Inverse functions (exercise)  (Page 2/3)

First combination :

$$\text{If}f\left(x\right)=1\text{is true, then}f\left(x\right)=1.$$

$$\text{If}f\left(y\right)\ne 1\text{is false, then}f\left(y\right)=1.$$

$$\text{If}f\left(z\right)\ne 2\text{is false, then}f\left(z\right)=2.$$

The function is clearly not one-one and hence not a bijection as f(x) = f(y).

Second combination :

$$\text{If}f\left(x\right)=1\text{is false, then}f\left(x\right)=2\text{or}3.$$

$$\text{If}f\left(y\right)\ne 1\text{is true, then}f\left(y\right)=2\text{or}3.$$

$$\text{If}f\left(z\right)\ne 2\text{is false, then}f\left(z\right)=2.$$

As f(z) = 2, f(x) = f(y) = 3. Again, the function is not one-one and hence not a bijection.

Third combination :

$$\text{If}f\left(x\right)=1\text{is false, then}f\left(x\right)=2\text{or}3.$$

$$\text{If}f\left(y\right)\ne 1\text{is false, then}f\left(y\right)=1.$$

$$\text{If}f\left(z\right)\ne 2\text{is true, then}f\left(z\right)=1\text{or}3.$$

As f(y) = 1, f(z) = 3 and f(x) = 2. In this case, we see that image and pre-image are related distinctly. The function is one-one and onto. Hence function is bijection for this combination. Also,

$$f^{-1}\left(1\right)=y$$

Problem 4: A function $f:R\to R$ is given by :

$$f\left(x\right)=x\left|x\right|$$

Find $f^{-1}\left(x\right)$ .

Solution :

Statement of the problem : We first need to see that the given function is bijection. If the function is bijection, then we find the rule of corresponding inverse function following the algorithm given in the beginning of the module.

By definition of modulus function,

$$f\left(x\right)=xX-x=-x^2-\infty <x\le 0$$

$$\Rightarrow f\left(x\right)=xXx=x^{2}0\le x<\infty$$

We take the derivative of the function to check whether the function is an injection :

$$\Rightarrow f\prime \left(x\right)=-2x-\infty <x\le 0$$

$$\Rightarrow f\prime \left(x\right)=2x0\le x<\infty$$

We see that derivative of function is non-negative number for all values of “x”

$$\Rightarrow f\prime \left(x\right)\ge 0$$

The equality holds only at a single point x = 0. Therefore, we conclude that function is an increasing function for all real values of “x”. It means that given function is one-one function in the domain of “R”. Now, we need to check whether function is onto function or not? For this, we find the range of the function. If range is “R”, then range is equal to co-domain, which is also “R”. For finding range of the function, we interpret the function, when “x” tends to become negative infinity or positive infinity (this is equivalent of taking limit).

$$\underset{x\to -\infty }{\overset{}{\lim }}f\left(x\right)=\underset{x\to -\infty }{\overset{}{\lim }}-x^2=-\infty$$

and

$$\underset{x\to \infty }{\overset{}{\lim }}f\left(x\right)=\underset{x\to \infty }{\overset{}{\lim }}{x}^2=\infty$$

The range of the function, therefore, is set of real numbers, “R”. Thus, we conclude that given function is an onto function.

The given function is both one-one and onto function and hence it is a bijection.

In order to find inverse function, we first solve the equation for “x” as :

For the interval, $-\infty <x\le 0$

$$\Rightarrow y=-x^2\Rightarrow x^2=-y\Rightarrow x=\pm \sqrt{\left(-y\right)}$$

But, we see that value of “x” is non-positive in the specified interval. Hence,

$$\Rightarrow x=-\sqrt{\left(-y\right)}$$

For the interval, $\Rightarrow 0\le x<\infty$

$$\Rightarrow y=x^2\Rightarrow x^2=y\Rightarrow x=\pm \sqrt{\left(y\right)}$$

But, we see that value of “x” is non-negative in the specified interval. Hence,

$$\Rightarrow x=\sqrt{\left(y\right)}$$

Substituting “x” by “ $f^{-1}\left(x\right)$ ” and “y” by “x”, we have the rule of inverse function as :

$$f^{-1}\left(x\right)=-\sqrt{\left(-x\right)}-\infty <x\le 0$$

and

$$f^{-1}\left(x\right)=\sqrt{\left(x\right)}0\le x<\infty$$

Problem 5: A quadratic function is given as :

$$f\left(x\right)=x^2+x+1$$

Is the function invertible? If not, then find the intervals in which it is invertible. Also find corresponding inverse functions.

Solution :

Statement of the problem : The given function is a continuous function valid for all values of “x”. We need to analyze function to determine whether function is bijection.

Let " $x_1$ " and " $x_2$ " be two values. Then,

$$x_1^2+x_1+1=x_2^2+x_2+1\Rightarrow x_1^2=x_2^2\Rightarrow x_1=\pm x_2$$

It means function is not one-one, but many one function. For determining whether function is onto function, we investigate the nature of its derivative,