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

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

Its root, when equated to zero, is -1/2. Its sign scheme is shown in the figure. The function value at "x = -1/2" corresponds to least value of the function.

$$\Rightarrow f\left(-\frac{1}{2}\right)={\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)+1=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$$

From the figure it is clear that function is strictly decreasing in the interval (-∞,-1/2] and strictly increasing in the interval [-1/2, ∞). Further, we also observe from the graph as shown in the figure below that function is one-one in the individual interval.

Now, for determining inverse function, we solve the function for "x". Here :

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

$$\Rightarrow x^2+x+1-y=0$$

Solving for “x”, we have :

$$\Rightarrow x=\frac{-1\pm \sqrt{\left(4y-3\right)}}{2}$$

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

For interval $\Rightarrow (-\infty ,-\frac{1}{2}]$

$$\Rightarrow f^{-1}\left(x\right)=\frac{-1-\sqrt{\left(4x-3\right)}}{2}$$

For interval $[-\frac{1}{2},\infty )$

$$\Rightarrow f^{-1}\left(x\right)=\frac{-1+\sqrt{\left(4x-3\right)}}{2}$$

Problem 6: An onto function is given as :

$$f\left(x\right)={\log }_a\{x+\sqrt{\left(x^2+1\right)}\},a>\mathrm{0,}a\ne 1$$

Is the function invertible? If invertible, then find $f^{-1}\left(x\right)$ .

Solution :

Statement of the problem : The given function is an onto function. This will be bijection i.e. invertible, if function is strictly monotonic (increasing or decreasing) so that function is one-one as well.

We see that $\sqrt{\left(x^2+1\right)}\ge 0$ for all values of “x”. Thus, argument of the logarithmic function is positive for all real values of “x”. It means that domain of the given function is “R”. In order to determine monotonic nature of the function, we differentiate given function in relevant intervals. Before differentiating, we need to convert the base from “a” to “e” as :

$$f\left(x\right)=\log e\{x+\sqrt{\left(x^2+1\right)}\}X{\log }_{a}e$$

Differentiating with respect to independent variable,

$$\Rightarrow f\prime \left(x\right)=\frac{{\log }_{a}e}{\{x+\sqrt{\left(x^2+1\right)}\}}X[1+\frac{2x}{\{x+\sqrt{\left(x^2+1\right)}\}}]$$

$$\Rightarrow f\prime \left(x\right)=\frac{{\log }_{a}e}{2\sqrt{\left(x^2+1\right)}}$$

Again $\sqrt{\left(x^2+1\right)}\ge 1$ i.e. a positive number. Hence, sign of f’(x) is same as that of $${\log }_{a}e$$ .

For $0<a<1$

See the graph. Here x = e = 2.718281828.

$$\Rightarrow {\log }_{a}e<0\Rightarrow f\prime \left(x\right)<0$$

The function is a decreasing function in the domain i.e. “R”

For $a>1$

See the graph. Here x = e = 2.718281828.

$$\Rightarrow {\log }_{a}e>0\Rightarrow f\prime \left(x\right)>0$$

The function is an increasing function in the domain i.e. “R”

We see that function is either increasing or decreasing for all real values of “x”. Thus, we conclude that function is strictly monotonic and function is on-one. This, in turn, means that function is bijection, which signifies that inverse of the function exists.

In order to find the inverse function, we write the logarithmic function in the equivalent exponential function as :

$$y=f\left(x\right)={\log }_a\{x+\sqrt{\left(x^2+1\right)}\}$$

$$\Rightarrow a^y=x+\sqrt{\left(x^2+1\right)}$$

Also,

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

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

$$\Rightarrow a^{-y}=\sqrt{\left(x^2+1\right)}-x$$

In order to solve for “x”, we subtract two equations :

$$\Rightarrow a^y-a^{-y}=x+\sqrt{\left(x^2+1\right)}-\sqrt{\left(x^2+1\right)}+x=2x$$

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

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

$$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}\left(a^x-a^{-x}\right)$$

Problem 7: A function $f:\left[\mathrm{1,}\infty \right)\to \left[\mathrm{1,}\infty \right)$ is given by :

$$f\left(x\right)=2^{x\left(x-1\right)}$$

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

Solution :

Statement of the problem : In order to determine the nature of given function to be a bijection, we need to check whether the function is both one-one and onto?

To check for one-one function, we determine the derivative of the function as :

$$f\left(x\right)=2^{x\left(x-1\right)}$$

Taking logarithm on both sides,

$$\log {}_e\left(f\left(x\right)\right)=x\left(x-1\right)\log {}_{e}2$$

Taking derivative on either side of the equation, we have :

$$\frac{f\prime \left(x\right)}{f\left(x\right)}=\left(2x-1\right){\log }_{e}2\Rightarrow f\prime \left(x\right)=f\left(x\right)\left(2x-1\right){\log }_{e}2$$

Now, “x” lie in the interval [1,∞ ). Hence, f(x) is a positive number. Also, ${\log }_{e}2$ is a positive number. Therefore,

$$f\prime \left(x\right)=\text{positive number}X\left(2x-1\right)$$

The root of the expression of f’(x), when equated to zero, is “0.5”. The derivative is positive in the interval between “1/2” and infinity. However, the domain of function begins at x = 1. We, therefore, conclude that the function is increasing in the interval given by [1,∞). The least value of the function is given as :

$$\Rightarrow f\left(1\right)=2^{x\left(x-1\right)}=2^0=1$$

We interpret the function, when “x” tends to become positive infinity (this is equivalent of taking limit).

$$\underset{x\to \infty }{\overset{}{\lim }}f\left(x\right)=\underset{x\to \infty }{\overset{}{\lim }}{2}^{x\left(x-1\right)}=2^{\infty X\left(\infty -1\right)}=2^{\infty X\infty }=2^{\infty }=\infty$$

Thus, range of the given function is [1, ∞). Clearly,

Range = Co-domain = [1, ∞)

Therefore, function is onto and hence bijection. It means that function is invertible. Now, we solve the function for “x” for finding inverse function. Taking log on the base of “2” on either side of the equation,

$$\Rightarrow \log {}_{2}y=\log {}_{2}x\left(x-1\right)=x\left(x-1\right)=x^2-x$$

$$\Rightarrow x^2-x-\log {}_{2}y=0$$

$$\Rightarrow x=\frac{1\pm \sqrt{\left(1+4X1X{\log }_{2}y\right)}}{2}=\frac{1\pm \sqrt{\left(1+4{\log }_{2}y\right)}}{2}$$

But, we know that domain of the function is x>= 1. Also, $\sqrt{\left(1+4{\log }_{2}y\right)}\ge 1$ . It means that only positive sign in the expression is valid.

$$\Rightarrow x=\frac{1+\sqrt{\left(1+4{\log }_{2}y\right)}}{2}$$

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

$$\Rightarrow f^{-1}\left(x\right)=\frac{1+\sqrt{\left(1+4{\log }_{2}x\right)}}{2}$$