Question

Let $f:D \to \mathbb{R}$, where $D$ is the domain of $f$ . Find the inverse of $f$, if it exists
A) $f\left( x \right) = 1 - {2^{ - x}}$
B) $f\left( x \right) = {\left( {4 - {{\left( {x - 7} \right)}^3}} \right)^{\dfrac{1}{5}}}$
C) $f\left( x \right) = \ln \left( {x + \sqrt {1 + {x^2}} } \right)$

Answer 1

The inverse of the function ${f^{ - 1}}$ exists only if the function $f\left( x \right)$ is one-one and onto.
One-one function: When each element of set B is mapped to only one element of set A, i.e., each object in set A has a unique image in set B, then the function is called one-one function.
onto functions. If for functions $f:{\text{A}} \to {\text{B}}$ the co-domain set of B is also the range for the function, then the function is called an onto function.

Complete step-by-step answer:
Step 1: Solve (1)
$f\left( x \right) = 1 - {2^{ - x}}$
On differentiating both sides.
$f'\left( x \right) = + {2^{ - x}}\ln 2 > 0$
$\Rightarrow f\left( x \right)$ is an increasing function.
Domain $= \mathbb{R}$ , where is a set of real numbers.
$\Rightarrow f\left( x \right)$ is a one-one function.
$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } 1 - {2^{ - x}} = - \infty$ $\left( {\because {2^\infty } = \infty } \right)$
$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } 1 - {2^{ - x}} = 1$ $\left( {\because {2^{ - \infty }} = 0} \right)$
$\therefore$ Range $= \left( { - \infty ,1} \right)$
But the codomain is $\mathbb{R}$ . Therefore function (1) is not onto. Hence, $f\left( x \right)$ is not invertible. Thus ${f^{ - 1}}$ i.e. inverse function does not exist.
Step 1: Solve (2)
$f\left( x \right) = {\left( {4 - {{\left( {x - 7} \right)}^3}} \right)^{\dfrac{1}{5}}}$
On differentiating both sides.
$f'\left( x \right) = \dfrac{1}{5}{\left( {4 - {{\left( {x - 7} \right)}^3}} \right)^{\dfrac{1}{5} - 1}} \times \left( { - 3} \right){\left( {x - 7} \right)^2} \\\ \Rightarrow \dfrac{{ - 3{{\left( {x - 7} \right)}^2}}}{{\dfrac{1}{5}{{\left( {4 - {{\left( {x - 7} \right)}^3}} \right)}^{\dfrac{4}{5}}}}} \\\$
For $f'\left( x \right)$ to be defined, the denominator should not be zero.
$4 - {\left( {x - 7} \right)^3} \ne 0 \\\ \Rightarrow {\left( {x - 7} \right)^3} \ne 4 \\\ \Rightarrow x - 7 \ne {4^{\dfrac{1}{3}}} \\\ \Rightarrow x \ne 7 + {4^{\dfrac{1}{3}}} \\\$
Domain, D = \mathbb{R} - \left\\{ {7 + {4^{\dfrac{1}{3}}}} \right\\} , where is a set of real numbers.
For $x \in D,{\text{ }}f'\left( x \right) > 0$
$\Rightarrow f\left( x \right)$ is a one-one function.
The range $= \mathbb{R} \Rightarrow f\left( x \right)$ is an onto function.
Thus $f\left( x \right)$ is an invertible function.
Let $f\left( x \right) = y$

$$y = {\left( {4 - {{\left( {x - 7} \right)}^3}} \right)^{\dfrac{1}{5}}} \\\ \Rightarrow {y^5} = 4 - {\left( {x - 7} \right)^3} \\\ \Rightarrow {\left( {x - 7} \right)^3} = 4 - {y^5} \\\ \Rightarrow x - 7 = {\left( {4 - {y^5}} \right)^{\dfrac{1}{3}}} \\\ \Rightarrow x = 7 + {\left( {4 - {y^5}} \right)^{\dfrac{1}{3}}} \\\$$

Interchange $x \leftrightarrow y$
$\Rightarrow y = 7 + {\left( {4 - {x^5}} \right)^{\dfrac{1}{3}}}$
The inverse of the function, ${f^{ - 1}}\left( x \right) = 7 + {\left( {4 - {x^5}} \right)^{\dfrac{1}{3}}}$
Step 3: Solve (3)
$f\left( x \right) = \ln \left( {x + \sqrt {1 + {x^2}} } \right)$
For domain: $x + \sqrt {1 + {x^2}} > 0,{\text{ }}\forall x \in \mathbb{R}$
Therefore, Domain $= \mathbb{R}$ , where is a set of real numbers.
Differentiating the given function on both sides.

$$f'\left( x \right) = \dfrac{{1 + \dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }} \\\ \Rightarrow \dfrac{{\dfrac{{x + \sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}} }}}}{{x + \sqrt {1 + {x^2}} }} \\\ \Rightarrow \dfrac{1}{{\sqrt {1 + {x^2}} }} \\\$$

$f'\left( x \right) > 0$
Therefore, the function is one-one.
Range $= \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \ln \left( {x + \sqrt {1 + {x^2}} } \right)$
Let $L = \mathop {\lim }\limits_{x \to - \infty } \ln \left( {x + \sqrt {1 + {x^2}} } \right)$
Put $x = - h$
$L = \mathop {\lim }\limits_{h \to \infty } \ln \left( { - h + \sqrt {1 + {{\left( { - h} \right)}^2}} } \right) \\\ \Rightarrow \mathop {\lim }\limits_{h \to \infty } \ln \left( { - h + \sqrt {1 + {{\left( h \right)}^2}} \times \dfrac{{\sqrt {1 + {{\left( h \right)}^2}} + h}}{{\sqrt {1 + {{\left( h \right)}^2}} + h}}} \right) \\\ \Rightarrow \mathop {\lim }\limits_{h \to \infty } \ln \left( {\dfrac{1}{{\sqrt {1 + {{\left( h \right)}^2}} + h}}} \right) \\\ \Rightarrow \mathop {\lim }\limits_{h \to \infty } - \ln \left( {\sqrt {1 + {{\left( h \right)}^2}} + h} \right) \\\ \Rightarrow - \infty \\\$
$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \ln \left( {x + \sqrt {1 + {x^2}} } \right) = \infty$
Therefore, Range $= \mathbb{R}$
Hence, the function is onto.
Thus, the function is invertible.
Let $f\left( x \right) = y$
$y = \ln \left( {x + \sqrt {1 + {x^2}} } \right) \\\ \Rightarrow {e^y} = \left( {x + \sqrt {1 + {x^2}} } \right) \\\ \Rightarrow {e^y} - x = \sqrt {1 + {x^2}} \\\$
Squaring both sides and expanding using the identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\Rightarrow {\left( {{e^y} - x} \right)^2} = {\left( {\sqrt {1 + {x^2}} } \right)^2} \\\ \Rightarrow {e^{2y}} + {x^2} - 2x{e^y} = 1 + {x^2} \\\ \Rightarrow {e^{2y}} - 2x{e^y} = 1 \\\ \Rightarrow {e^{2y}} - 1 = 2x{e^y} \\\ \Rightarrow x = \dfrac{{{e^{2y}} - 1}}{{2{e^y}}} \\\$
Interchange $x \leftrightarrow y$
$\Rightarrow y = \dfrac{{{e^{2x}} - 1}}{{2{e^x}}}$

The inverse of the function, ${f^{ - 1}}\left( x \right) = \dfrac{{{e^{2x}} - 1}}{{2{e^x}}}$

Note: The composition of functions is also useful in checking the invertible function, hence finding the inverse of the function.
A function $f:X \to Y$ is defined to be invertible, if there exists a function $g:Y \to X$ such that $gof = {I_x}$and $fog = {I_y}$ . The function $g$ is called the inverse of $f$ and is denoted by ${f^{ - 1}}$ .