Inverse Function Questions in English

(C) Given $f(x) = x + \tan x$ and $g(x)$ is the inverse of $f(x)$,so $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$f'(g(x)) \cdot g'(x) = 1$
$g'(x) = \frac{1}{f'(g(x))}$.
Since $f'(x) = 1 + \sec^2 x$,we have $f'(g(x)) = 1 + \sec^2(g(x))$.
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$,we get:
$f'(g(x)) = 1 + (1 + \tan^2(g(x))) = 2 + \tan^2(g(x))$.
Thus,$g'(x) = \frac{1}{2 + \tan^2(g(x))}$.
From the definition $f(g(x)) = x$,we have:
$g(x) + \tan(g(x)) = x$
$\tan(g(x)) = x - g(x)$.
Squaring both sides:
$\tan^2(g(x)) = (x - g(x))^2 = (g(x) - x)^2$.
Substituting this into the expression for $g'(x)$:
$g'(x) = \frac{1}{2 + (g(x) - x)^2}$.

(D) function $f: A \to B$ is invertible if and only if it is a bijection (both one-to-one and onto).
Option $A$: $y = e^x$ is a strictly increasing function from $R$ to $R^+$,hence it is a bijection and invertible.
Option $B$: $y = \log|x|$ for $x \in R^+$,$y = \log x$,which is a strictly increasing function from $R^+$ to $R$,hence it is a bijection and invertible.
Option $C$: $y = \sin^3x$ is a strictly increasing function on $\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]$ mapping to $[-1, 1]$,hence it is a bijection and invertible.
Option $D$: $y = e^{[x]}$ where $[x]$ is the greatest integer function. For any $x \in [0, 1)$,$[x] = 0$,so $y = e^0 = 1$. Since the function takes the same value for all $x$ in the interval $[0, 1)$,it is not one-to-one (many-to-one). Therefore,it is not a bijection and its inverse cannot be defined.

(A) Given $f(x) = \log_a(x + \sqrt{x^2 + 1})$.
Let $y = f(x)$,so $y = \log_a(x + \sqrt{x^2 + 1})$.
By the definition of logarithm,$a^y = x + \sqrt{x^2 + 1}$.
Now,consider $a^{-y} = \frac{1}{a^y} = \frac{1}{x + \sqrt{x^2 + 1}}$.
Rationalizing the denominator: $a^{-y} = \frac{\sqrt{x^2 + 1} - x}{(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x)} = \frac{\sqrt{x^2 + 1} - x}{(x^2 + 1) - x^2} = \sqrt{x^2 + 1} - x$.
We have two equations:
$1) \sqrt{x^2 + 1} + x = a^y$
$2) \sqrt{x^2 + 1} - x = a^{-y}$
Subtracting equation $(2)$ from equation $(1)$:
$(\sqrt{x^2 + 1} + x) - (\sqrt{x^2 + 1} - x) = a^y - a^{-y}$
$2x = a^y - a^{-y}$
$x = \frac{a^y - a^{-y}}{2}$.
Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{a^y - a^{-y}}{2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{a^x - a^{-x}}{2}$.

(C) To find $f^{-1}(17)$,we set $f(x) = 17$:
$x^2 + 1 = 17$
$x^2 = 16$
$x = \pm 4$.
Thus,$f^{-1}(17) = \{4, -4\}$.
To find $f^{-1}(-3)$,we set $f(x) = -3$:
$x^2 + 1 = -3$
$x^2 = -4$.
Since the square of a real number cannot be negative,there is no real value of $x$ that satisfies this equation.
Thus,$f^{-1}(-3) = \phi$ (the empty set).
Therefore,the values are $\{4, -4\}$ and $\phi$.

(C) Given the domain $x \in (4, 6)$.
For $x \in (4, 6)$,we have $\frac{x}{2} \in (2, 3)$.
The greatest integer function $[\frac{x}{2}]$ for any value in $(2, 3)$ is equal to $2$.
Therefore,the function becomes $f(x) = x + 2$.
To find the inverse,let $y = f(x) = x + 2$.
Solving for $x$,we get $x = y - 2$.
Thus,$f^{-1}(x) = x - 2$.

(D) To check if the function is invertible,we need to show it is both one-to-one (injective) and onto (surjective).
$1$. One-to-one: Let $f(x_1) = f(x_2)$.
$\frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2}$
$(x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2)$
$x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$
$-2x_1 - x_2 = -2x_2 - x_1$
$x_2 = x_1$. Thus,$f$ is one-to-one.
$2$. Onto: Let $y = \frac{x - 1}{x - 2}$.
$y(x - 2) = x - 1$
$yx - 2y = x - 1$
$yx - x = 2y - 1$
$x(y - 1) = 2y - 1$
$x = \frac{2y - 1}{y - 1}$.
Since for every $y \in R - \{1\}$,there exists an $x \in R - \{2\}$,the function is onto.
Since $f$ is both one-to-one and onto,it is invertible,and $f^{-1}(y) = \frac{2y - 1}{y - 1}$.

(C) Given $f(x) = x^2 - x + 5$ for $x > \frac{1}{2}$.
To find $g'(7)$,we use the formula for the derivative of an inverse function: $g'(y) = \frac{1}{f'(x)}$,where $y = f(x)$.
First,find $x$ such that $f(x) = 7$:
$x^2 - x + 5 = 7 \implies x^2 - x - 2 = 0$.
$(x - 2)(x + 1) = 0$.
Since $x > \frac{1}{2}$,we have $x = 2$.
Now,find $f'(x)$:
$f'(x) = \frac{d}{dx}(x^2 - x + 5) = 2x - 1$.
At $x = 2$,$f'(2) = 2(2) - 1 = 3$.
Therefore,$g'(7) = \frac{1}{f'(2)} = \frac{1}{3}$.

(B) Let $y = \frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}$ for $x \ne 0$.
To find $f^{-1}(x)$,we solve $y = \frac{x - 1}{x + 2}$ for $x$ in terms of $y$:
$y(x + 2) = x - 1$
$xy + 2y = x - 1$
$xy - x = -2y - 1$
$x(y - 1) = -(2y + 1)$
$x = \frac{2y + 1}{1 - y}$
Thus,$f^{-1}(x) = \frac{2x + 1}{1 - x}$.
Now,differentiate $f^{-1}(x)$ with respect to $x$ using the quotient rule:
$\frac{d}{dx}[f^{-1}(x)] = \frac{d}{dx}\left( \frac{2x + 1}{1 - x} \right)$
$= \frac{(1 - x)(2) - (2x + 1)(-1)}{(1 - x)^2}$
$= \frac{2 - 2x + 2x + 1}{(1 - x)^2}$
$= \frac{3}{(1 - x)^2}$

(C) Let $y = f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}$.
Multiply numerator and denominator by $8^{2x}$:
$y = \frac{8^{4x} - 1}{8^{4x} + 1}$.
Now,solve for $x$ in terms of $y$:
$y(8^{4x} + 1) = 8^{4x} - 1$
$y \cdot 8^{4x} + y = 8^{4x} - 1$
$1 + y = 8^{4x}(1 - y)$
$8^{4x} = \frac{1 + y}{1 - y}$.
Taking $\log_{8}$ on both sides:
$4x = \log_{8} \left(\frac{1 + y}{1 - y}\right)$.
Using the change of base formula $\log_{8} A = \frac{\ln A}{\ln 8} = (\log_{8} e) \ln A$:
$4x = (\log_{8} e) \ln \left(\frac{1 + y}{1 - y}\right)$.
$x = \frac{1}{4} (\log_{8} e) \ln \left(\frac{1 + y}{1 - y}\right)$.
Replacing $y$ with $x$,the inverse function is $f^{-1}(x) = \frac{1}{4} (\log_{8} e) \ln \left(\frac{1 + x}{1 - x}\right)$.

(A) Define a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g(a) = 1$,$g(b) = 2$,and $g(c) = 3$.
Now,consider the composite function $g \circ f: X \rightarrow X$:
$(g \circ f)(1) = g(f(1)) = g(a) = 1$
$(g \circ f)(2) = g(f(2)) = g(b) = 2$
$(g \circ f)(3) = g(f(3)) = g(c) = 3$
Since $(g \circ f)(x) = x$ for all $x \in X$,$g \circ f = I_X$.
Next,consider the composite function $f \circ g: Y \rightarrow Y$:
$(f \circ g)(a) = f(g(a)) = f(1) = a$
$(f \circ g)(b) = f(g(b)) = f(2) = b$
$(f \circ g)(c) = f(g(c)) = f(3) = c$
Since $(f \circ g)(y) = y$ for all $y \in Y$,$f \circ g = I_Y$.

(D) To show that $f$ is invertible,we must find a function $g: Y \rightarrow N$ such that $g \circ f = I_N$ and $f \circ g = I_Y$.
Consider an arbitrary element $y \in Y$. By the definition of $Y$,$y = 4x + 3$ for some $x \in N$.
Solving for $x$,we get $x = \frac{y-3}{4}$.
Define $g: Y \rightarrow N$ by $g(y) = \frac{y-3}{4}$.
Now,calculate the composition $g \circ f(x)$:
$g(f(x)) = g(4x+3) = \frac{(4x+3)-3}{4} = \frac{4x}{4} = x$.
Thus,$g \circ f = I_N$.
Next,calculate the composition $f \circ g(y)$:
$f(g(y)) = f\left(\frac{y-3}{4}\right) = 4\left(\frac{y-3}{4}\right) + 3 = (y-3) + 3 = y$.
Thus,$f \circ g = I_Y$.
Since $g \circ f = I_N$ and $f \circ g = I_Y$,the function $f$ is invertible and its inverse is $f^{-1}(y) = \frac{y-3}{4}$.

(N/A) To show that $f$ is invertible,we need to find a function $g: Y \rightarrow N$ such that $g \circ f = I_{N}$ and $f \circ g = I_{Y}$.
An arbitrary element $y$ in $Y$ is of the form $n^{2}$ for some $n \in N$.
This implies that $n = \sqrt{y}$.
We define a function $g: Y \rightarrow N$ by $g(y) = \sqrt{y}$.
Now,calculate the compositions:
$(g \circ f)(n) = g(f(n)) = g(n^{2}) = \sqrt{n^{2}} = n = I_{N}(n)$.
$(f \circ g)(y) = f(g(y)) = f(\sqrt{y}) = (\sqrt{y})^{2} = y = I_{Y}(y)$.
Since $g \circ f = I_{N}$ and $f \circ g = I_{Y}$,the function $f$ is invertible.
The inverse function is $f^{-1}(y) = \sqrt{y}$ for all $y \in Y$.

(N/A) Let $y$ be an arbitrary element of the range $S$. Then $y = 4x^{2} + 12x + 15$ for some $x \in N$.
This can be written as $y = (2x + 3)^{2} + 6$.
Solving for $x$,we get $(2x + 3)^{2} = y - 6$,which implies $2x + 3 = \sqrt{y - 6}$ (since $x \in N$,$2x + 3 > 0$).
Thus,$x = \frac{\sqrt{y - 6} - 3}{2}$.
Define $g: S \rightarrow N$ by $g(y) = \frac{\sqrt{y - 6} - 3}{2}$.
Now,$g(f(x)) = g((2x + 3)^{2} + 6) = \frac{\sqrt{(2x + 3)^{2} + 6 - 6} - 3}{2} = \frac{(2x + 3) - 3}{2} = x$.
Also,$f(g(y)) = f\left(\frac{\sqrt{y - 6} - 3}{2}\right) = \left(2\left(\frac{\sqrt{y - 6} - 3}{2}\right) + 3\right)^{2} + 6 = (\sqrt{y - 6} - 3 + 3)^{2} + 6 = (\sqrt{y - 6})^{2} + 6 = y - 6 + 6 = y$.
Since $g \circ f = I_{N}$ and $f \circ g = I_{S}$,$f$ is invertible and $f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}$.

By definition,$f$ and $g$ are bijective functions because they are one-to-one and onto.
$f^{-1}: \{a, b, c\} \rightarrow \{1, 2, 3\}$ is defined as $f^{-1}(a)=1, f^{-1}(b)=2, f^{-1}(c)=3$.
$g^{-1}: \{\text{apple, ball, cat}\} \rightarrow \{a, b, c\}$ is defined as $g^{-1}(\text{apple})=a, g^{-1}(\text{ball})=b, g^{-1}(\text{cat})=c$.
Since $f^{-1} \circ f = I_{\{1, 2, 3\}}$ and $f \circ f^{-1} = I_{\{a, b, c\}}$,$f$ is invertible.
Since $g^{-1} \circ g = I_{\{a, b, c\}}$ and $g \circ g^{-1} = I_{\{\text{apple, ball, cat}\}}$,$g$ is invertible.
$g \circ f: \{1, 2, 3\} \rightarrow \{\text{apple, ball, cat}\}$ is defined as $(g \circ f)(1)=\text{apple}, (g \circ f)(2)=\text{ball}, (g \circ f)(3)=\text{cat}$.
$(g \circ f)^{-1}: \{\text{apple, ball, cat}\} \rightarrow \{1, 2, 3\}$ is defined as $(g \circ f)^{-1}(\text{apple})=1, (g \circ f)^{-1}(\text{ball})=2, (g \circ f)^{-1}(\text{cat})=3$.
Now,$(f^{-1} \circ g^{-1})(\text{apple}) = f^{-1}(g^{-1}(\text{apple})) = f^{-1}(a) = 1 = (g \circ f)^{-1}(\text{apple})$.
$(f^{-1} \circ g^{-1})(\text{ball}) = f^{-1}(g^{-1}(\text{ball})) = f^{-1}(b) = 2 = (g \circ f)^{-1}(\text{ball})$.
$(f^{-1} \circ g^{-1})(\text{cat}) = f^{-1}(g^{-1}(\text{cat})) = f^{-1}(c) = 3 = (g \circ f)^{-1}(\text{cat})$.
Thus,$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$.

(B) function $f: S \rightarrow S$ is invertible if and only if it is a bijection (both one-one and onto).
Given $f = \{(1, 1), (2, 2), (3, 3)\}$,we observe that each element in the domain $S$ is mapped to a unique element in the codomain $S$,and every element in the codomain has a pre-image.
Thus,$f$ is one-one and onto.
Since $f$ is a bijection,it is invertible.
The inverse function $f^{-1}$ is obtained by interchanging the elements of the ordered pairs $(x, y)$ to $(y, x)$.
Therefore,$f^{-1} = \{(1, 1), (2, 2), (3, 3)\} = f$.

(B) function $f: A \to B$ is invertible if and only if it is both one-one (injective) and onto (surjective).
Given $f = \{(1,2), (2,1), (3,1)\}$,we observe the values:
$f(1) = 2$
$f(2) = 1$
$f(3) = 1$
Since $f(2) = f(3) = 1$ but $2 \neq 3$,the function is not one-one.
Because the function is not one-one,it cannot be invertible.
Therefore,the correct answer is that $f$ is not invertible.

(B) function $f: A \to B$ is invertible if and only if it is a bijection (one-one and onto).
Given $f = \{(1,2), (2,3), (3,1)\}$.
The inverse function $f^{-1}: B \to A$ is obtained by interchanging the elements of the ordered pairs $(x, y)$ to $(y, x)$.
Thus,$f^{-1} = \{(2,1), (3,2), (1,3)\}$.
Rearranging the pairs,we get $f^{-1} = \{(1,3), (2,1), (3,2)\}$.
Therefore,the correct option is $B$.

(N/A) Given $f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3}.$
$(f \circ f)(x) = f(f(x)) = f\left(\frac{4x+3}{6x-4}\right)$
$= \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$
$= \frac{\frac{16x+12 + 18x-12}{6x-4}}{\frac{24x+18 - 24x+16}{6x-4}}$
$= \frac{34x}{34} = x.$
Since $(f \circ f)(x) = x = I(x),$ the function $f$ is its own inverse.
Therefore,the inverse of $f$ is $f$ itself,i.e.,$f^{-1}(x) = f(x) = \frac{4x+3}{6x-4}.$

(D) The function $f: \{1,2,3,4\} \rightarrow \{10\}$ is defined as $f = \{(1,10), (2,10), (3,10), (4,10)\}$.
From the given definition of $f$,we observe that $f(1) = f(2) = f(3) = f(4) = 10$.
Since multiple elements in the domain map to the same element in the codomain,$f$ is a many-one function.
Therefore,$f$ is not one-one (injective).
$A$ function has an inverse if and only if it is bijective (both one-one and onto).
Since $f$ is not one-one,it does not have an inverse.

(NO) function $g$ has an inverse if and only if it is a bijection (both one-one and onto).
Given $g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$.
We observe that $g(5) = 4$ and $g(7) = 4$.
Since two distinct elements in the domain,$5$ and $7$,map to the same element $4$ in the codomain,the function $g$ is not one-one (it is many-one).
Because $g$ is not one-one,it is not a bijection.
Therefore,the function $g$ does not have an inverse.

(A) function $h$ has an inverse if and only if it is a bijection (both one-one and onto).
$1$. Check for one-one: The function $h$ maps distinct elements of the domain $\{2, 3, 4, 5\}$ to distinct elements in the codomain $\{7, 9, 11, 13\}$. Specifically,$h(2)=7, h(3)=9, h(4)=11, h(5)=13$. Since no two elements in the domain have the same image,$h$ is one-one.
$2$. Check for onto: The range of $h$ is $\{7, 9, 11, 13\}$,which is equal to the codomain. Thus,every element in the codomain has a pre-image in the domain. Hence,$h$ is onto.
Since $h$ is both one-one and onto,it is a bijection. Therefore,the function $h$ has an inverse.

(N/A) For one-one,let $f(x_1) = f(x_2)$.
$\Rightarrow \frac{x_1}{x_1+2} = \frac{x_2}{x_2+2}$
$\Rightarrow x_1(x_2+2) = x_2(x_1+2)$
$\Rightarrow x_1x_2 + 2x_1 = x_1x_2 + 2x_2$
$\Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$.
Thus,$f$ is one-one.
For the inverse,let $y = f(x) = \frac{x}{x+2}$.
$y(x+2) = x \Rightarrow xy + 2y = x \Rightarrow 2y = x(1-y) \Rightarrow x = \frac{2y}{1-y}$.
Since $f$ is onto its range,the inverse function $f^{-1}: \text{Range } f \rightarrow [-1,1]$ is defined by $f^{-1}(y) = \frac{2y}{1-y}$.

(N/A) To show that $f$ is invertible,we must prove that $f$ is both one-one and onto.
$1$. One-one:
Let $f(x_1) = f(x_2)$.
$4x_1 + 3 = 4x_2 + 3$
$4x_1 = 4x_2$
$x_1 = x_2$.
Thus,$f$ is one-one.
$2$. Onto:
For any $y \in R$,let $y = 4x + 3$.
Then $4x = y - 3$,which gives $x = \frac{y-3}{4}$.
Since $y \in R$,$x = \frac{y-3}{4} \in R$.
For every $y \in R$,there exists $x = \frac{y-3}{4}$ such that $f(x) = 4(\frac{y-3}{4}) + 3 = y - 3 + 3 = y$.
Thus,$f$ is onto.
Since $f$ is both one-one and onto,it is invertible.
To find $f^{-1}$,we use the relation $f(x) = y \implies x = f^{-1}(y)$.
From $y = 4x + 3$,we have $x = \frac{y-3}{4}$.
Therefore,$f^{-1}(y) = \frac{y-3}{4}$,or $f^{-1}(x) = \frac{x-3}{4}$.

(A) $f: R_{+} \rightarrow [4, \infty)$ is given as $f(x) = x^{2} + 4$.
For one-one:
Let $f(x) = f(y)$.
$\Rightarrow x^{2} + 4 = y^{2} + 4$.
$\Rightarrow x^{2} = y^{2}$.
$\Rightarrow x = y$ (since $x, y \in R_{+}$).
Therefore,$f$ is a one-one function.
For onto:
For $y \in [4, \infty)$,let $y = x^{2} + 4$.
$\Rightarrow x^{2} = y - 4 \geq 0$ (as $y \geq 4$).
$\Rightarrow x = \sqrt{y - 4} \geq 0$.
Therefore,for any $y \in [4, \infty)$,there exists $x = \sqrt{y - 4} \in R_{+}$,such that $f(x) = f(\sqrt{y - 4}) = (\sqrt{y - 4})^{2} + 4 = y - 4 + 4 = y$.
Therefore,$f$ is onto.
Thus,$f$ is one-one and onto,and therefore $f^{-1}$ exists.
Let us define $g: [4, \infty) \rightarrow R_{+}$ by $g(y) = \sqrt{y - 4}$.
Now,
$(g \circ f)(x) = g(f(x)) = g(x^{2} + 4) = \sqrt{(x^{2} + 4) - 4} = \sqrt{x^{2}} = x$.
And
$(f \circ g)(y) = f(g(y)) = f(\sqrt{y - 4}) = (\sqrt{y - 4})^{2} + 4 = y - 4 + 4 = y$.
Therefore,$g \circ f = I_{R_{+}}$ and $f \circ g = I_{[4, \infty)}$.
Hence,$f$ is invertible and the inverse of $f$ is given by $f^{-1}(y) = g(y) = \sqrt{y - 4}$.

(A) $f: R_{+} \rightarrow [-5, \infty)$ is given by $f(x) = 9x^{2} + 6x - 5$.
Let $y$ be an arbitrary element of $[-5, \infty)$.
Set $y = 9x^{2} + 6x - 5$.
$y = (3x + 1)^{2} - 1 - 5 = (3x + 1)^{2} - 6$.
$y + 6 = (3x + 1)^{2}$.
Since $x \in R_{+}$,$x > 0$,so $3x + 1 > 1$. Thus,$3x + 1 = \sqrt{y+6}$.
$x = \frac{\sqrt{y+6}-1}{3}$.
Define $g: [-5, \infty) \rightarrow R_{+}$ by $g(y) = \frac{\sqrt{y+6}-1}{3}$.
$(g \circ f)(x) = g(f(x)) = g((3x+1)^{2}-6) = \frac{\sqrt{(3x+1)^{2}-6+6}-1}{3} = \frac{3x+1-1}{3} = x$.
$(f \circ g)(y) = f(g(y)) = 9\left(\frac{\sqrt{y+6}-1}{3}\right)^{2} + 6\left(\frac{\sqrt{y+6}-1}{3}\right) - 5 = (\sqrt{y+6}-1)^{2} + 2(\sqrt{y+6}-1) - 5 = (y+6 - 2\sqrt{y+6} + 1) + 2\sqrt{y+6} - 2 - 5 = y + 7 - 7 = y$.
Since $g \circ f = I_{R_{+}}$ and $f \circ g = I_{[-5, \infty)}$,$f$ is invertible and $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

Let $f: X \rightarrow Y$ be an invertible function.
Suppose $f$ has two inverses,$g_{1}$ and $g_{2}$,where $g_{1}: Y \rightarrow X$ and $g_{2}: Y \rightarrow X$.
By the definition of an inverse function,for any $y \in Y$,we have:
$f \circ g_{1}(y) = I_{Y}(y) = y$
$f \circ g_{2}(y) = I_{Y}(y) = y$
Therefore,$f(g_{1}(y)) = f(g_{2}(y))$ for all $y \in Y$.
Since $f$ is an invertible function,it must be a bijection (both one-one and onto).
Because $f$ is one-one,$f(x_{1}) = f(x_{2}) \Rightarrow x_{1} = x_{2}$.
Applying this to $f(g_{1}(y)) = f(g_{2}(y))$,we get $g_{1}(y) = g_{2}(y)$ for all $y \in Y$.
Thus,$g_{1} = g_{2}$.
Hence,the inverse of an invertible function $f$ is unique.

(N/A) The function $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ is defined as $f(1) = a, f(2) = b, f(3) = c$.
To find $f^{-1}$,we define a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g(a) = 1, g(b) = 2, g(c) = 3$.
We check the compositions:
$(f \circ g)(a) = f(g(a)) = f(1) = a$
$(f \circ g)(b) = f(g(b)) = f(2) = b$
$(f \circ g)(c) = f(g(c)) = f(3) = c$
Thus,$f \circ g = I_Y$,where $Y = \{a, b, c\}$.
$(g \circ f)(1) = g(f(1)) = g(a) = 1$
$(g \circ f)(2) = g(f(2)) = g(b) = 2$
$(g \circ f)(3) = g(f(3)) = g(c) = 3$
Thus,$g \circ f = I_X$,where $X = \{1, 2, 3\}$.
Since $f \circ g = I_Y$ and $g \circ f = I_X$,the inverse of $f$ exists and $f^{-1} = g$.
Therefore,$f^{-1}: \{a, b, c\} \rightarrow \{1, 2, 3\}$ is defined by $f^{-1}(a) = 1, f^{-1}(b) = 2, f^{-1}(c) = 3$.
Now,to find $(f^{-1})^{-1}$,we find the inverse of $g$. Let $h: \{1, 2, 3\} \rightarrow \{a, b, c\}$ be defined as $h(1) = a, h(2) = b, h(3) = c$.
We check the compositions:
$(g \circ h)(1) = g(h(1)) = g(a) = 1$
$(g \circ h)(2) = g(h(2)) = g(b) = 2$
$(g \circ h)(3) = g(h(3)) = g(c) = 3$
Thus,$g \circ h = I_X$.
$(h \circ g)(a) = h(g(a)) = h(1) = a$
$(h \circ g)(b) = h(g(b)) = h(2) = b$
$(h \circ g)(c) = h(g(c)) = h(3) = c$
Thus,$h \circ g = I_Y$.
Since $g \circ h = I_X$ and $h \circ g = I_Y$,the inverse of $g$ exists and $g^{-1} = h$. Since $g = f^{-1}$,we have $(f^{-1})^{-1} = h$. Since $h = f$,it follows that $(f^{-1})^{-1} = f$.

(A) Let $f : X \rightarrow Y$ be an invertible function.
By definition,a function $f$ is invertible if there exists a function $g : Y \rightarrow X$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $I_X$ and $I_Y$ are identity functions on $X$ and $Y$ respectively.
In this case,$g = f^{-1}$.
Substituting $g = f^{-1}$ into the conditions,we have:
$f^{-1} \circ f = I_X$ and $f \circ f^{-1} = I_Y$.
Now,consider the function $f^{-1} : Y \rightarrow X$. For $f^{-1}$ to be invertible,there must exist a function $h : X \rightarrow Y$ such that $h \circ f^{-1} = I_Y$ and $f^{-1} \circ h = I_X$.
From the conditions $f \circ f^{-1} = I_Y$ and $f^{-1} \circ f = I_X$,we can see that $f$ acts as the function $h$.
Therefore,$f$ is the inverse of $f^{-1}$,which means $\left(f^{-1}\right)^{-1} = f$.

(A) Given $f(x) = \frac{4x}{3x+4}$.
To find the inverse,set $y = f(x) = \frac{4x}{3x+4}$.
$y(3x+4) = 4x$
$3xy + 4y = 4x$
$4y = 4x - 3xy$
$4y = x(4-3y)$
$x = \frac{4y}{4-3y}$.
Thus,the inverse function $g(y) = \frac{4y}{4-3y}$.

(B) Given the function $y = 5^{\log x}$.
To find the inverse,we interchange $x$ and $y$:
$x = 5^{\log y}$.
Using the property of logarithms $a^{\log b} = b^{\log a}$,we can rewrite the expression:
$x = y^{\log 5}$.
Thus,the inverse function is $f^{-1}(y) = y^{\log 5}$.

(A) Given $f(x) = 10x + 7$. For $f$ to be invertible,it must be one-one and onto.
$1$. One-one: Let $f(x_1) = f(x_2)$. Then $10x_1 + 7 = 10x_2 + 7$,which implies $10x_1 = 10x_2$,so $x_1 = x_2$. Thus,$f$ is one-one.
$2$. Onto: Let $y = 10x + 7$. Solving for $x$,we get $x = \frac{y-7}{10}$. Since for every $y \in R$,there exists $x = \frac{y-7}{10} \in R$,$f$ is onto.
Since $f$ is bijective,it is invertible. Let $g(y) = f^{-1}(y)$.
$g(y) = \frac{y-7}{10}$.
Verification:
$(g \circ f)(x) = g(f(x)) = g(10x + 7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x = I_R(x)$.
$(f \circ g)(y) = f(g(y)) = f\left(\frac{y-7}{10}\right) = 10\left(\frac{y-7}{10}\right) + 7 = y - 7 + 7 = y = I_R(y)$.
Thus,$g(x) = \frac{x-7}{10}$.

(A) It is given that $f: W \rightarrow W$ is defined as:
$f(n) = \begin{cases} n-1 & \text{if } n \text{ is odd} \\ n+1 & \text{if } n \text{ is even} \end{cases}$
For one-one:
Let $f(n) = f(m)$.
If $n$ is odd and $m$ is even,then $n-1 = m+1 \Rightarrow n-m = 2$. This is impossible as the difference between an odd and an even number is always odd. Similarly,$n$ being even and $m$ being odd is impossible.
Therefore,$n$ and $m$ must have the same parity.
If both are odd,$n-1 = m-1 \Rightarrow n = m$.
If both are even,$n+1 = m+1 \Rightarrow n = m$.
Thus,$f$ is one-one.
For onto:
Any odd number $2r+1$ in the codomain is the image of $2r+2$ (which is even) or $2r$ (which is odd). Specifically,$f(2r+2) = 2r+1$ and $f(2r) = 2r+1$ is not the case here; rather,$f(2r+2) = 2r+3$ and $f(2r+1) = 2r$. Every $m \in W$ has a preimage. Thus,$f$ is onto.
Since $f$ is one-one and onto,it is invertible.
Let $g: W \rightarrow W$ be defined as $g(m) = \begin{cases} m+1 & \text{if } m \text{ is even} \\ m-1 & \text{if } m \text{ is odd} \end{cases}$.
We observe that $f(f(n)) = n$ for all $n \in W$.
If $n$ is odd,$f(n) = n-1$ (even),so $f(f(n)) = f(n-1) = (n-1)+1 = n$.
If $n$ is even,$f(n) = n+1$ (odd),so $f(f(n)) = f(n+1) = (n+1)-1 = n$.
Thus,$f \circ f = I_W$,which implies $f^{-1} = f$.

(A) Given sets are $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$.
The function $F : S \rightarrow T$ is defined as $F = \{(a, 3), (b, 2), (c, 1)\}$.
This implies $F(a) = 3$,$F(b) = 2$,and $F(c) = 1$.
Since $F$ is a one-to-one and onto (bijective) function,its inverse $F^{-1} : T \rightarrow S$ exists.
The inverse function is obtained by reversing the ordered pairs of $F$.
Therefore,$F^{-1} = \{(3, a), (2, b), (1, c)\}$.

(D) Given $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$.
The function $F : S \rightarrow T$ is defined as $F = \{(a, 2), (b, 1), (c, 1)\}$.
$A$ function is invertible if and only if it is both one-one (injective) and onto (surjective).
Here,$F(b) = 1$ and $F(c) = 1$. Since $F(b) = F(c)$ but $b \neq c$,the function $F$ is not one-one.
Because $F$ is not one-one,it is not a bijection.
Therefore,$F$ is not invertible,which means $F^{-1}$ does not exist.

(C) Given $f(x) = \frac{x-2}{x-3}$. Let $y = \frac{x-2}{x-3}$.
$y(x-3) = x-2 \implies yx - 3y = x - 2 \implies x(y-1) = 3y-2 \implies x = \frac{3y-2}{y-1}$.
Thus,$f^{-1}(x) = \frac{3x-2}{x-1}$.
Given $g(x) = 2x-3$. Let $y = 2x-3$.
$y+3 = 2x \implies x = \frac{y+3}{2}$.
Thus,$g^{-1}(x) = \frac{x+3}{2}$.
We are given $f^{-1}(x) + g^{-1}(x) = \frac{13}{2}$.
$\frac{3x-2}{x-1} + \frac{x+3}{2} = \frac{13}{2}$.
Multiply by $2(x-1)$: $2(3x-2) + (x+3)(x-1) = 13(x-1)$.
$6x - 4 + x^2 + 2x - 3 = 13x - 13$.
$x^2 + 8x - 7 = 13x - 13$.
$x^2 - 5x + 6 = 0$.
$(x-2)(x-3) = 0$.
The roots are $x = 2$ and $x = 3$. However,$f^{-1}(x)$ is defined for $x \in R - \{1\}$. Since $x=3$ is in the domain,we check the equation: $f^{-1}(3) + g^{-1}(3) = \frac{3(3)-2}{3-1} + \frac{3+3}{2} = \frac{7}{2} + 3 = \frac{13}{2}$. Both values are valid.
The sum of the values of $x$ is $2 + 3 = 5$.

(B) Given $f(x) = \frac{5x + 3}{6x - \alpha}$.
For $(f \circ f)(x) = x$,the function $f$ must be its own inverse,i.e.,$f(x) = f^{-1}(x)$.
Let $y = f(x) = \frac{5x + 3}{6x - \alpha}$.
Then $y(6x - \alpha) = 5x + 3$.
$6xy - \alpha y = 5x + 3$.
$6xy - 5x = \alpha y + 3$.
$x(6y - 5) = \alpha y + 3$.
$x = \frac{\alpha y + 3}{6y - 5}$.
Thus,$f^{-1}(x) = \frac{\alpha x + 3}{6x - 5}$.
Since $f(x) = f^{-1}(x)$,we have $\frac{5x + 3}{6x - \alpha} = \frac{\alpha x + 3}{6x - 5}$.
Comparing the coefficients,we get $\alpha = 5$.

(A) Given $(g \circ f)(x) = x$. By the chain rule,$g'(f(x)) \cdot f'(x) = 1$.
We need to find $g'(2)$. Let $f(x) = 2$.
$x^5 + 2e^{x/4} = 2$. By inspection,$x = 0$ satisfies this equation $(0^5 + 2e^0 = 2)$.
Thus,$g'(f(0)) \cdot f'(0) = 1$,which implies $g'(2) = \frac{1}{f'(0)}$.
Now,$f'(x) = 5x^4 + 2 \cdot \frac{1}{4} e^{x/4} = 5x^4 + \frac{1}{2} e^{x/4}$.
At $x = 0$,$f'(0) = 5(0)^4 + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.
Therefore,$g'(2) = \frac{1}{1/2} = 2$.
The value of $8g'(2) = 8 \times 2 = 16$.

(D) Given $f(x)=x^3+3x+2$. Since $g(f(x))=x$,$g$ is the inverse of $f$.
Differentiating $g(f(x))=x$ with respect to $x$,we get $g'(f(x)) \cdot f'(x)=1$.
For $g'(2)$,we set $f(x)=2 \implies x^3+3x+2=2 \implies x^3+3x=0 \implies x(x^2+3)=0$,so $x=0$.
Thus $g'(2) \cdot f'(0)=1$. Since $f'(x)=3x^2+3$,$f'(0)=3$.
So $g'(2) = \frac{1}{3}$. Option $A$ is incorrect.
Given $h(g(g(x)))=x$. Since $g(f(x))=x$,$g(g(f(f(x))))=x$,which implies $g(g(x))=f(f(x))$.
Thus $h(f(f(x)))=x$. This means $h$ is the inverse of $f(f(x))$.
However,the problem states $h(g(g(x)))=x$. Since $g(g(x))=f^{-1}(f^{-1}(x))$,we have $h(f^{-1}(f^{-1}(x)))=x$.
Let $f^{-1}(x)=y$,then $f(y)=x$. The equation becomes $h(f^{-1}(y))=f(y)$.
Let $f^{-1}(y)=z$,then $y=f(z)$. So $h(z)=f(f(z))$.
Thus $h(x)=f(f(x))$.
Now,$h'(x)=f'(f(x)) \cdot f'(x)$.
$h'(1)=f'(f(1)) \cdot f'(1)$. Since $f(1)=1^3+3(1)+2=6$ and $f'(x)=3x^2+3$,$f'(1)=6$ and $f'(6)=3(6^2)+3=111$.
$h'(1)=111 \times 6=666$. Option $B$ is correct.
$h(0)=f(f(0))=f(2)=2^3+3(2)+2=16$. Option $C$ is correct.
$h(g(3))=f(f(g(3)))=f(3)=3^3+3(3)+2=38$. Option $D$ is incorrect.
Therefore,the correct options are $B$ and $C$.

(C) Given $f(x) = x^3 + e^{x/2}$.
We need to find $g^{\prime}(1)$ where $g = f^{-1}$.
First,find $x$ such that $f(x) = 1$.
$x^3 + e^{x/2} = 1$.
By inspection,if $x = 0$,then $f(0) = 0^3 + e^0 = 0 + 1 = 1$.
Thus,$f(0) = 1$,which implies $g(1) = 0$.
Using the formula for the derivative of an inverse function: $g^{\prime}(y) = \frac{1}{f^{\prime}(x)}$ where $y = f(x)$.
Here $y = 1$,so $x = 0$.
$f^{\prime}(x) = \frac{d}{dx}(x^3 + e^{x/2}) = 3x^2 + \frac{1}{2}e^{x/2}$.
At $x = 0$,$f^{\prime}(0) = 3(0)^2 + \frac{1}{2}e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.
Therefore,$g^{\prime}(1) = \frac{1}{f^{\prime}(0)} = \frac{1}{1/2} = 2$.

(B) Given $e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \dots (i)$
At $x = 0$,$e^0 f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \implies f(0) = 2$.
Since $f(0) = 2$,it follows that $f^{-1}(2) = 0$.
We know that $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$.
For $y = 2$,$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)}$.
Differentiating equation $(i)$ with respect to $x$ using the product rule and the Fundamental Theorem of Calculus:
$-e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1}$.
At $x = 0$:
$-e^0 f(0) + e^0 f'(0) = \sqrt{0^4 + 1}$
$-1(2) + 1(f'(0)) = 1$
$-2 + f'(0) = 1 \implies f'(0) = 3$.
Therefore,$(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{3}$.

(C) Given $f(x) = \frac{b-x}{1-bx}$.
To find the inverse,let $y = f(x) = \frac{b-x}{1-bx}$.
$y(1-bx) = b-x \Rightarrow y - bxy = b - x \Rightarrow x(1-by) = b-y \Rightarrow x = \frac{b-y}{1-by}$.
Thus,$f^{-1}(y) = \frac{b-y}{1-by}$,which implies $f(x) = f^{-1}(x)$,so $f = f^{-1}$.
Now,$f^{\prime}(x) = \frac{(1-bx)(-1) - (b-x)(-b)}{(1-bx)^2} = \frac{-1+bx+b^2-bx}{(1-bx)^2} = \frac{b^2-1}{(1-bx)^2}$.
$f^{\prime}(0) = \frac{b^2-1}{(1-0)^2} = b^2-1$.
$f^{\prime}(b) = \frac{b^2-1}{(1-b^2)^2} = \frac{b^2-1}{(b^2-1)^2} = \frac{1}{b^2-1}$.
Therefore,$f^{\prime}(b) = \frac{1}{f^{\prime}(0)}$.

(B) Let $h(x) = f(g^{-1}(x))$. We need to find $h'(2) = f'(g^{-1}(2)) \cdot (g^{-1})'(2)$.
First,find $g^{-1}(2)$. Since $g(0) = \frac{4}{1 + e^0} = \frac{4}{2} = 2$,we have $g^{-1}(2) = 0$.
Next,find $f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$. Thus,$f'(0) = \frac{2}{4} = 0.5$.
Now,find $(g^{-1})'(2)$. We know $(g^{-1})'(g(x)) = \frac{1}{g'(x)}$.
$g'(x) = \frac{4 \cdot (-1) \cdot e^{-2x} \cdot (-2)}{(1 + e^{-2x})^2} = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$.
At $x = 0$,$g'(0) = \frac{8(1)}{(1 + 1)^2} = \frac{8}{4} = 2$.
Therefore,$(g^{-1})'(2) = \frac{1}{g'(0)} = \frac{1}{2} = 0.5$.
Finally,$h'(2) = f'(0) \cdot (g^{-1})'(2) = 0.5 \cdot 0.5 = 0.25$.

(B) Given $f(x) = \frac{x^2-x}{x^2+2x}$. For $x \neq 0$,we can simplify this as $f(x) = \frac{x-1}{x+2}$.
Let $y = \frac{x-1}{x+2}$.
To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:
$y(x+2) = x-1$
$yx + 2y = x - 1$
$yx - x = -1 - 2y$
$x(y-1) = -(1+2y)$
$x = \frac{2y+1}{1-y}$.
Thus,$f^{-1}(x) = \frac{2x+1}{1-x}$.
Now,we differentiate $f^{-1}(x)$ with respect to $x$ using the quotient rule:
$\frac{d}{dx}(f^{-1}(x)) = \frac{d}{dx}\left(\frac{2x+1}{1-x}\right) = \frac{(1-x)(2) - (2x+1)(-1)}{(1-x)^2}$
$= \frac{2 - 2x + 2x + 1}{(1-x)^2} = \frac{3}{(1-x)^2}$.
Evaluating at $x = 2$:
$\frac{d}{dx}(f^{-1}(x))|_{x=2} = \frac{3}{(1-2)^2} = \frac{3}{(-1)^2} = 3$.

(B) Given that $g(x)$ is the inverse of the function $f(x)$,we have $g(x) = f^{-1}(x)$.
This implies $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$
Therefore,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$ ... $(i)$
Given $f^{\prime}(x) = \frac{1}{1+x^3}$,we substitute $x$ with $g(x)$ to get:
$f^{\prime}(g(x)) = \frac{1}{1+(g(x))^3}$ ... $(ii)$
Substituting $(ii)$ into $(i)$,we get:
$g^{\prime}(x) = \frac{1}{\frac{1}{1+(g(x))^3}} = 1+(g(x))^3$.

(A) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
We are given $f^{\prime}(x) = \frac{1}{1+x^4}$.
Substituting $g(x)$ for $x$ in the expression for $f^{\prime}(x)$,we get $f^{\prime}(g(x)) = \frac{1}{1+[g(x)]^4}$.
Substituting this into the differentiated equation: $\frac{1}{1+[g(x)]^4} \cdot g^{\prime}(x) = 1$.
Therefore,$g^{\prime}(x) = 1+[g(x)]^4$.

(C) Given $f(x) = x^3 + e^{\frac{x}{2}}$.
First,find the value of $x$ such that $f(x) = 1$.
$x^3 + e^{\frac{x}{2}} = 1$.
By inspection,if $x = 0$,then $f(0) = 0^3 + e^0 = 0 + 1 = 1$.
Thus,$f(0) = 1$,which implies $g(1) = 0$.
We know that $g(f(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$g^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$.
To find $g^{\prime}(1)$,we set $x = 0$:
$g^{\prime}(f(0)) \cdot f^{\prime}(0) = 1 \Rightarrow g^{\prime}(1) \cdot f^{\prime}(0) = 1$.
Now,calculate $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^2 + \frac{1}{2} e^{\frac{x}{2}}$.
$f^{\prime}(0) = 3(0)^2 + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.
Therefore,$g^{\prime}(1) = \frac{1}{f^{\prime}(0)} = \frac{1}{1/2} = 2$.

(A) Step $1$: Find $f^{-1}(x)$. Let $y = \frac{x-3}{x-2}$. Then $y(x-2) = x-3$,so $xy - 2y = x - 3$. Thus $x(y-1) = 2y-3$,which gives $x = \frac{2y-3}{y-1}$. So,$f^{-1}(x) = \frac{2x-3}{x-1}$.
Step $2$: Find $g^{-1}(x)$. Let $y = 3x-2$. Then $3x = y+2$,so $x = \frac{y+2}{3}$. Thus,$g^{-1}(x) = \frac{x+2}{3}$.
Step $3$: Solve the equation $f^{-1}(x) + g^{-1}(x) = \frac{19}{6}$.
$\frac{2x-3}{x-1} + \frac{x+2}{3} = \frac{19}{6}$.
Multiply by $6(x-1)$ to clear denominators: $6(2x-3) + 2(x-1)(x+2) = 19(x-1)$.
$12x - 18 + 2(x^2 + x - 2) = 19x - 19$.
$12x - 18 + 2x^2 + 2x - 4 = 19x - 19$.
$2x^2 + 14x - 22 = 19x - 19$.
$2x^2 - 5x - 3 = 0$.
Step $4$: Solve the quadratic equation $2x^2 - 5x - 3 = 0$.
$(2x+1)(x-3) = 0$.
The roots are $x = -\frac{1}{2}$ and $x = 3$.
Step $5$: The sum of the values of $x$ is $-\frac{1}{2} + 3 = \frac{5}{2}$.

(A) Given $f(x) = x + \frac{1}{x}$.
Let $y = f(x) = x + \frac{1}{x}$.
Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.
Using the quadratic formula to solve for $x$:
$x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.
Since the domain is $x \in [1, \infty)$,we must have $x \ge 1$.
If we take $x = \frac{y - \sqrt{y^2 - 4}}{2}$,then for $y=2$,$x=1$,but for $y > 2$,this value would be less than $1$.
Thus,we choose the positive root: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.
Therefore,$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

(A) Let $f(x) = y$,which implies $x = f^{-1}(y)$.
Given $y = \frac{2x - 3}{3x - 4}$.
Multiply both sides by $(3x - 4)$:
$y(3x - 4) = 2x - 3$
$3xy - 4y = 2x - 3$
Rearrange the terms to isolate $x$:
$3xy - 2x = 4y - 3$
$x(3y - 2) = 4y - 3$
$x = \frac{4y - 3}{3y - 2}$
Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{4y - 3}{3y - 2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{4x - 3}{3x - 2}$.