Inverse Function Questions in English

(A) Let $y = f(x) = \frac{a^x - a^{-x}}{a^x + a^{-x}}$.
Multiply the numerator and denominator by $a^x$:
$y = \frac{a^{2x} - 1}{a^{2x} + 1}$.
Now,solve for $x$ in terms of $y$:
$y(a^{2x} + 1) = a^{2x} - 1$
$y \cdot a^{2x} + y = a^{2x} - 1$
$1 + y = a^{2x} - y \cdot a^{2x}$
$1 + y = a^{2x}(1 - y)$
$a^{2x} = \frac{1+y}{1-y}$.
Taking the logarithm base $a$ on both sides:
$2x = \log_a \left( \frac{1+y}{1-y} \right)$
$x = \frac{1}{2} \log_a \left( \frac{1+y}{1-y} \right)$.
Replacing $y$ with $x$ to find the inverse function:
$f^{-1}(x) = \frac{1}{2} \log_a \left( \frac{1+x}{1-x} \right)$.

(C) Given $f(x) = 7x + 8 = y$.
To find the inverse function $f^{-1}(y)$,we solve for $x$ in terms of $y$:
$7x = y - 8$
$x = \frac{y - 8}{7}$
Thus,$f^{-1}(y) = \frac{y - 8}{7}$,which implies $f^{-1}(x) = \frac{x - 8}{7}$.
Given $f^{-1}(12) = \frac{k}{7}$,we substitute $x = 12$ into the inverse function:
$f^{-1}(12) = \frac{12 - 8}{7} = \frac{4}{7}$.
Comparing $\frac{4}{7}$ with $\frac{k}{7}$,we get $k = 4$.

(B) To find the inverse function $f^{-1}(x)$,we set $f(x) = y$.
$y = \frac{4x}{5} + 3$
Subtract $3$ from both sides:
$y - 3 = \frac{4x}{5}$
Multiply both sides by $5$:
$5(y - 3) = 4x$
Divide by $4$:
$x = \frac{5(y - 3)}{4}$
Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{5(y - 3)}{4}$.
Replacing $y$ with $x$,we get:
$f^{-1}(x) = \frac{5(x - 3)}{4}$
Thus,the correct option is $B$.

(A) Let $y = f(x) = \frac{3x+2}{5x-3}$.
Then $y(5x - 3) = 3x + 2$.
Expanding this,we get $5xy - 3y = 3x + 2$.
Rearranging the terms to solve for $x$,we have $5xy - 3x = 3y + 2$.
Factoring out $x$,we get $x(5y - 3) = 3y + 2$.
Thus,$x = \frac{3y+2}{5y-3}$.
By definition,$f^{-1}(y) = x = \frac{3y+2}{5y-3}$.
Replacing $y$ with $x$,we obtain $f^{-1}(x) = \frac{3x+2}{5x-3}$.
Since $f^{-1}(x) = f(x)$,the correct option is $A$.

(A) Given that $g$ is the inverse of $f$,so $g(x) = f^{-1}(x)$.
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,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.
Given $f^{\prime}(x) = \frac{1}{1+x^{2}}$,we substitute $g(x)$ for $x$:
$f^{\prime}(g(x)) = \frac{1}{1+[g(x)]^{2}}$.
Substituting this into the expression for $g^{\prime}(x)$:
$g^{\prime}(x) = \frac{1}{\frac{1}{1+[g(x)]^{2}}} = 1 + [g(x)]^{2}$.

(B) Let $y = f(x) = x^{3} + 5$.
To find the inverse,we express $x$ in terms of $y$:
$y - 5 = x^{3}$
$x = (y - 5)^{1/3}$
Since $f^{-1}(y) = x$,we have $f^{-1}(y) = (y - 5)^{1/3}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = (x - 5)^{1/3}$.

(B) Given $f(x) = (x+1)^2 - 1$ for $x \geqslant -1$.
Since $f(x)$ is an increasing function for $x \geqslant -1$,the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$.
Setting $f(x) = x$:
$(x+1)^2 - 1 = x$
$x^2 + 2x + 1 - 1 = x$
$x^2 + x = 0$
$x(x+1) = 0$
This gives $x = 0$ or $x = -1$.
Both values satisfy the condition $x \geqslant -1$.
Thus,the set is $\{0, -1\}$.

(B) For a function $f: N \rightarrow N$ to have an inverse $f^{-1}$,the function must be a bijection (both one-one and onto).
Given $f(x) = x + 3$,where $N$ is the set of natural numbers.
For the function to be onto,the range must be equal to the codomain.
The range of $f(x) = x + 3$ for $x \in N$ is $\{4, 5, 6, \dots \}$.
The codomain is $N = \{1, 2, 3, 4, 5, 6, \dots \}$.
Since the range $\{4, 5, 6, \dots \} \neq N$,the function is not onto.
Therefore,$f^{-1}(x)$ does not exist.

(B) To find the inverse of the function $f(x) = \frac{3x+1}{5x-3}$,let $y = f(x)$.
So,$y = \frac{3x+1}{5x-3}$.
Cross-multiplying gives $y(5x-3) = 3x+1$,which simplifies to $5xy - 3y = 3x + 1$.
Rearranging the terms to solve for $x$: $5xy - 3x = 3y + 1$.
Factoring out $x$: $x(5y - 3) = 3y + 1$.
Thus,$x = \frac{3y+1}{5y-3}$.
By definition,$f^{-1}(y) = x = \frac{3y+1}{5y-3}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{3x+1}{5x-3}$.
Since $f^{-1}(x) = f(x)$,the correct option is $B$.

(A) Given $f(x) = 3x + 2$. Let $y = 3x + 2$,then $x = \frac{y - 2}{3}$.
Thus,$f^{-1}(y) = \frac{y - 2}{3}$,or $f^{-1}(x) = \frac{x - 2}{3}$.
We need to find $(g \circ f^{-1})(10) = g(f^{-1}(10))$.
First,calculate $f^{-1}(10) = \frac{10 - 2}{3} = \frac{8}{3}$.
Now,substitute this into $g(x) = 6x + 5$:
$g\left(\frac{8}{3}\right) = 6 \left(\frac{8}{3}\right) + 5 = 2(8) + 5 = 16 + 5 = 21$.
Therefore,the correct option is $A$.

(A) To find the inverse function $f^{-1}(x)$,let $y = f(x)$.
Given $y = 4x + 3$.
Now,solve for $x$ in terms of $y$:
$y - 3 = 4x$
$x = \frac{y - 3}{4}$.
Since $f^{-1}(y) = x$,we have $f^{-1}(y) = \frac{y - 3}{4}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x - 3}{4}$.
Therefore,the correct option is $A$.

(A) Given $f(x) = \sin 2x + \cos 2x$ and $g(x) = x^2 - 1$.
First,calculate the composite function $g(f(x))$:
$g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$
$g(f(x)) = (\sin^2 2x + \cos^2 2x + 2 \sin 2x \cos 2x) - 1$
Since $\sin^2 \theta + \cos^2 \theta = 1$ and $2 \sin \theta \cos \theta = \sin 2\theta$:
$g(f(x)) = (1 + \sin 4x) - 1 = \sin 4x$.
$A$ function is invertible if it is one-to-one (injective) and onto (surjective) in the given domain.
The function $y = \sin \theta$ is invertible in the interval $\theta \in \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
Here,$\theta = 4x$,so we set:
$\frac{-\pi}{2} \le 4x \le \frac{\pi}{2}$
Dividing by $4$:
$\frac{-\pi}{8} \le x \le \frac{\pi}{8}$.
Thus,$g(f(x))$ is invertible in the domain $x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$.

(B) Given $f(x) = \tan x$.
The inverse function $f^{-1}(y)$ is defined such that $f(f^{-1}(y)) = y$.
Here,we need to find $f^{-1}(1)$,so we set $f(x) = 1$.
$\tan x = 1$.
The general solution for $\tan x = \tan \alpha$ is $x = n \pi + \alpha$,where $n \in Z$.
Since $\tan(\frac{\pi}{4}) = 1$,the set of values for $x$ is $\{n \pi + \frac{\pi}{4} : n \in Z\}$.
Therefore,$f^{-1}(1) = \{n \pi + \frac{\pi}{4} : n \in Z\}$.

(A) Given the function $f(x) = 2x + 6$,where $f: R \rightarrow R$.
To find the inverse $f^{-1}(x)$,let $y = f(x)$.
So,$y = 2x + 6$.
Solving for $x$ in terms of $y$:
$2x = y - 6$
$x = \frac{y - 6}{2}$
$x = \frac{y}{2} - 3$.
By definition,$f^{-1}(y) = x$,so $f^{-1}(y) = \frac{y}{2} - 3$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x}{2} - 3$.

(A) Let $x_{1}, x_{2} \in R$.
For injectivity,let $f(x_{1}) = f(x_{2})$.
$2x_{1} + 3 = 2x_{2} + 3$
$2x_{1} = 2x_{2}$
$x_{1} = x_{2}$.
Hence,$f$ is injective.
For surjectivity,let $y \in \text{codomain } R$.
Let $y = f(x) = 2x + 3$.
$y - 3 = 2x$
$x = \frac{y-3}{2}$.
Since for every $y \in R$,there exists an $x = \frac{y-3}{2} \in R$,$f$ is surjective.
Since $f$ is both injective and surjective,$f^{-1}$ exists.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x-3}{2}$.

(A) Given the function $f(x) = x^{3}$.
To find $f^{-1}(8)$,we need to solve the equation $f(x) = 8$ for $x$.
$x^{3} = 8$
$x^{3} - 8 = 0$
$(x - 2)(x^{2} + 2x + 4) = 0$
This gives $x = 2$ as the real solution.
Since the domain of $f$ is $R$ (the set of real numbers),we only consider the real value.
Therefore,$f^{-1}(8) = \{2\}$.

(C) We have $f(x) = |x|$.
For a function to be invertible,it must be both one-one (injective) and onto (surjective).
Consider the values $f(1) = |1| = 1$ and $f(-1) = |-1| = 1$.
Since $f(1) = f(-1)$ but $1 \neq -1$,the function is not one-one.
Because the function is not one-one,it is not bijective.
Therefore,the inverse function $f^{-1}(x)$ does not exist.

(B) Given that $f(x)=(x+1)^{2}$ for $x \geq -1$.
Since $g(x)$ is the reflection of the graph of $f(x)$ in the line $y=x$,$g(x)$ is the inverse function of $f(x)$,denoted as $f^{-1}(x)$.
To find the inverse,let $y = (x+1)^{2}$.
Since $x \geq -1$,we have $x+1 \geq 0$,so taking the square root of both sides gives $\sqrt{y} = x+1$.
Solving for $x$,we get $x = \sqrt{y} - 1$.
By interchanging $x$ and $y$,we obtain $f^{-1}(x) = \sqrt{x} - 1$.
Thus,$g(x) = \sqrt{x} - 1$.

(B) For $f^{-1}(x)$ to exist,the function $f(x)$ must be a bijection (both one-to-one and onto).
Given $f(x) = 2 + |\sin^{-1} x|$.
The domain of $\sin^{-1} x$ is $[-1, 1]$.
For $x \in [-1, 0]$,$f(x) = 2 - \sin^{-1} x$. This is a strictly decreasing function.
For $x \in [0, 1]$,$f(x) = 2 + \sin^{-1} x$. This is a strictly increasing function.
Since the function decreases on $[-1, 0]$ and increases on $[0, 1]$,it is not one-to-one on the interval $[-1, 1]$.
Specifically,$f(x_1) = f(x_2)$ for some $x_1 \in [-1, 0)$ and $x_2 \in (0, 1]$.
Therefore,$f(x)$ is not invertible on its entire domain $[-1, 1]$.
However,if the question implies the set $A$ where the inverse exists locally or is defined,typically such functions are not invertible. Given the options provided,there is no valid interval where the function is globally invertible. If we assume the question implies the domain where the function is defined,it is $[-1, 1]$. But strictly speaking,$f^{-1}(x)$ does not exist for the whole domain.

(A) Given $f(x) = \sin x + \cos x$ and $g(x) = x^2 - 1$.
$g(f(x)) = g(\sin x + \cos x) = (\sin x + \cos x)^2 - 1$.
$= \sin^2 x + \cos^2 x + 2 \sin x \cos x - 1$.
$= 1 + \sin 2x - 1 = \sin 2x$.
For a function to be invertible,it must be one-one and onto in the given domain.
The function $h(x) = \sin 2x$ is one-one in the interval where the argument $2x$ lies within $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Thus,$-\frac{\pi}{2} \leq 2x \leq \frac{\pi}{2}$.
Dividing by $2$,we get $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$.
Therefore,the function is invertible in the interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$.

(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.
To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$.
Then $y+1 = (x+1)^2$. Since $x \geq -1$,$x+1 = \sqrt{y+1}$,so $x = \sqrt{y+1} - 1$.
Thus,$f^{-1}(x) = \sqrt{x+1} - 1$.
The equation $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ because $f$ is an increasing function for $x \geq -1$.
So,$(x+1)^2 - 1 = x$.
$x^2 + 2x + 1 - 1 = x$.
$x^2 + x = 0$.
$x(x+1) = 0$.
This gives $x = 0$ or $x = -1$.
Both values satisfy the condition $x \geq -1$.
Therefore,the set is $\{0, -1\}$.

(A) Given that $g(x)$ is the inverse of the function $f(x)$,we have $f(g(x)) = x$.
By differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
This implies $g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.
We are given $f^{\prime}(x) = \frac{1}{h(x)}$,which means $f^{\prime}(g(x)) = \frac{1}{h(g(x))}$.
Substituting this into the expression for $g^{\prime}(x)$:
$g^{\prime}(x) = \frac{1}{1 / h(g(x))} = h(g(x))$.
Thus,the correct option is $A$.

(C) To find the inverse of the function $f(x) = (x + 2)^2 - 2$ for $x \geq -2$,we set $y = f(x)$.
$y = (x + 2)^2 - 2$
Add $2$ to both sides:
$y + 2 = (x + 2)^2$
Since $x \geq -2$,we have $x + 2 \geq 0$. Taking the positive square root of both sides:
$\sqrt{y + 2} = x + 2$
Subtract $2$ from both sides to solve for $x$:
$x = \sqrt{y + 2} - 2$
By definition,$f^{-1}(y) = \sqrt{y + 2} - 2$. Replacing $y$ with $x$,we get:
$f^{-1}(x) = \sqrt{x + 2} - 2$

(A) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.
Multiply the numerator and denominator by $10^x$:
$y = \frac{10^{2x} - 1}{10^{2x} + 1}$
Now,solve for $x$ in terms of $y$:
$y(10^{2x} + 1) = 10^{2x} - 1$
$y \cdot 10^{2x} + y = 10^{2x} - 1$
$1 + y = 10^{2x} - y \cdot 10^{2x}$
$1 + y = 10^{2x}(1 - y)$
$10^{2x} = \frac{1 + y}{1 - y}$
Taking $\log_{10}$ on both sides:
$2x = \log_{10} \left( \frac{1 + y}{1 - y} \right)$
$x = \frac{1}{2} \log_{10} \left( \frac{1 + y}{1 - y} \right)$
Replacing $y$ with $x$ to find the inverse function $f^{-1}(x)$:
$f^{-1}(x) = \frac{1}{2} \log_{10} \left( \frac{1 + x}{1 - x} \right)$

(C) Given the function $f:(5,10) \rightarrow (7,12)$ defined by $f(x) = x + 2\left[\frac{x}{5}\right]$.
For $x \in (5, 10)$,the value of $\frac{x}{5}$ lies in the interval $(1, 2)$.
Therefore,the greatest integer function $\left[\frac{x}{5}\right] = 1$ for all $x \in (5, 10)$.
Substituting this into the function definition,we get $f(x) = x + 2(1) = x + 2$.
Since $f(x) = x + 2$ is a linear function,it is strictly increasing and therefore one-to-one (injective).
To check if it is onto (surjective),we find the range: as $x$ varies from $5$ to $10$,$f(x)$ varies from $5+2=7$ to $10+2=12$. Thus,the range is $(7, 12)$,which matches the codomain.
Since $f$ is bijective,$f^{-1}(x)$ exists.
Let $y = x + 2$,then $x = y - 2$.
Thus,$f^{-1}(x) = x - 2$.

(B) Given,$f(x) = 5x - 3$ and $g(x) = x^2 + 3$.
To find $f^{-1}(x)$,let $y = f(x) = 5x - 3$.
Then $y + 3 = 5x$,which implies $x = \frac{y + 3}{5}$.
Thus,$f^{-1}(y) = \frac{y + 3}{5}$,so $f^{-1}(x) = \frac{x + 3}{5}$.
Now,we need to calculate $g \circ f^{-1}(3) = g(f^{-1}(3))$.
First,find $f^{-1}(3) = \frac{3 + 3}{5} = \frac{6}{5}$.
Then,$g(\frac{6}{5}) = (\frac{6}{5})^2 + 3 = \frac{36}{25} + 3$.
Calculating the sum: $\frac{36 + 75}{25} = \frac{111}{25}$.

(C) Given functions are $f(x) = 3x - 4$ and $g(x) = 3x + 2$.
To find $f^{-1}(y)$,let $f(x) = y$.
$3x - 4 = y \implies 3x = y + 4 \implies x = \frac{y + 4}{3}$.
Thus,$f^{-1}(y) = \frac{y + 4}{3}$.
Now,$f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3$.
To find $g^{-1}(z)$,let $g(x) = z$.
$3x + 2 = z \implies 3x = z - 2 \implies x = \frac{z - 2}{3}$.
Thus,$g^{-1}(z) = \frac{z - 2}{3}$.
Finally,$g^{-1}(f^{-1}(5)) = g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3}$.

(B) 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 $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
Therefore,$f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$.
Given $g(x) = x + \tan x$,we find $g^{\prime}(x) = 1 + \sec^2 x$.
Substituting this into the expression,we get $f^{\prime}(g(x)) = \frac{1}{1 + \sec^2 x}$.
Let $y = g(x)$,then $x = f(y)$.
Substituting $x = f(y)$ into the equation,we get $f^{\prime}(y) = \frac{1}{1 + \sec^2(f(y))}$.
Replacing $y$ with $x$,we obtain $f^{\prime}(x) = \frac{1}{1 + \sec^2(f(x))}$.

(B) Given $(f \circ g)(x) = x$,$g(x)$ is the inverse function of $f(x)$.
Let $g(1 + (2n - 1) \pi) = x_n$. Then $f(x_n) = 1 + (2n - 1) \pi$.
Substituting $f(x) = 2x + \cos x + \sin^2 x$:
$2x_n + \cos x_n + \sin^2 x_n = 1 + (2n - 1) \pi$
$2x_n + \cos x_n + 1 - \cos^2 x_n = 1 + (2n - 1) \pi$
$2x_n + \cos x_n - \cos^2 x_n = (2n - 1) \pi$.
For $x_n = (2n - 1) \frac{\pi}{2}$,we have $\cos x_n = 0$.
Substituting this into the equation: $2((2n - 1) \frac{\pi}{2}) + 0 - 0 = (2n - 1) \pi$,which holds true.
Thus,$g(1 + (2n - 1) \pi) = (2n - 1) \frac{\pi}{2}$.
Now,$\sum_{n=1}^{99} g(1 + (2n - 1) \pi) = \sum_{n=1}^{99} (2n - 1) \frac{\pi}{2} = \frac{\pi}{2} \sum_{n=1}^{99} (2n - 1)$.
The sum of the first $99$ odd numbers is $99^2$.
Therefore,the sum is $(99)^2 \frac{\pi}{2}$.

(D) Given the function $y = \frac{10^x - 10^{-x}}{10^x + 10^{-x}} + 1$.
Subtract $1$ from both sides: $y - 1 = \frac{10^x - 10^{-x}}{10^x + 10^{-x}}$.
Let $u = 10^x$. Then $10^{-x} = \frac{1}{u}$.
So,$y - 1 = \frac{u - 1/u}{u + 1/u} = \frac{u^2 - 1}{u^2 + 1}$.
Let $Y = y - 1$. Then $Y(u^2 + 1) = u^2 - 1$.
$Yu^2 + Y = u^2 - 1 \implies Y + 1 = u^2(1 - Y)$.
$u^2 = \frac{1 + Y}{1 - Y} = \frac{1 + (y - 1)}{1 - (y - 1)} = \frac{y}{2 - y}$.
Since $u = 10^x$,we have $10^{2x} = \frac{y}{2 - y}$.
Taking $\log_{10}$ on both sides: $2x = \log_{10} \left(\frac{y}{2 - y}\right)$.
Therefore,$x = \frac{1}{2} \log_{10} \left(\frac{y}{2 - y}\right)$.

(D) For two functions $f: A \rightarrow B$ and $g: B \rightarrow A$ to be inverses of each other,they must satisfy $g(f(x)) = x$ for all $x \in A$ and $f(g(x)) = x$ for all $x \in B$.
Given $f(x) = x^2$ and $g(x) = x^{1/2}$.
For $g(x) = \sqrt{x}$ to be defined,we must have $x \geq 0$ for all $x \in B$. Thus,$B = [0, \infty) = R \setminus R^{-}$.
For $f(g(x)) = (x^{1/2})^2 = x$,this holds for all $x \in B$.
For $g(f(x)) = (x^2)^{1/2} = |x|$,we require $|x| = x$,which implies $x \geq 0$. Thus,$A = [0, \infty) = R \setminus R^{-}$.
Therefore,$f(x)$ and $g(x)$ are inverse functions to each other when $A = B = R \setminus R^{-}$.

(B) Given,$f(x) = x + x^2 + x^3 + \ldots \infty$.
This is an infinite geometric series with first term $a = x$ and common ratio $r = x$.
Since the domain is $(-1, 1)$,$|x| < 1$,the sum is given by $f(x) = \frac{x}{1-x}$.
To find the range $B$,we set $y = \frac{x}{1-x}$.
$y(1-x) = x \Rightarrow y - xy = x \Rightarrow y = x(1+y) \Rightarrow x = \frac{y}{1+y}$.
Since $-1 < x < 1$,we have $-1 < \frac{y}{1+y} < 1$.
Case $1$: $\frac{y}{1+y} > -1 \Rightarrow \frac{y + 1 + y}{1+y} > 0 \Rightarrow \frac{2y+1}{1+y} > 0$.
Critical points are $y = -1$ and $y = -1/2$. Testing intervals,we get $y \in (-\infty, -1) \cup (-1/2, \infty)$.
Case $2$: $\frac{y}{1+y} < 1 \Rightarrow \frac{y - 1 - y}{1+y} < 0 \Rightarrow \frac{-1}{1+y} < 0 \Rightarrow \frac{1}{1+y} > 0 \Rightarrow y > -1$.
Taking the intersection of Case $1$ and Case $2$,we get $y \in (-1/2, \infty)$.
Thus,$B = (-1/2, \infty)$.

(D) Given $f(x) = x^2 + 3x - 3$.
Completing the square,we get $f(x) = (x + \frac{3}{2})^2 - \frac{9}{4} - 3 = (x + \frac{3}{2})^2 - \frac{21}{4}$.
For the inverse to exist,the function must be monotonic. The vertex of the parabola is at $x = -\frac{3}{2}$,so the function is strictly increasing on $[-\frac{3}{2}, \infty)$. Thus,$\alpha = -\frac{3}{2}$.
Let $y = (x + \frac{3}{2})^2 - \frac{21}{4}$. Then $x + \frac{3}{2} = \sqrt{y + \frac{21}{4}}$,so $g(x) = f^{-1}(x) = \sqrt{x + \frac{21}{4}} - \frac{3}{2}$.
We need to find $g'(x)$ at $x = \alpha + \frac{5}{2} = -\frac{3}{2} + \frac{5}{2} = 1$.
$g'(x) = \frac{d}{dx}(\sqrt{x + \frac{21}{4}} - \frac{3}{2}) = \frac{1}{2\sqrt{x + \frac{21}{4}}}$.
At $x = 1$,$g'(1) = \frac{1}{2\sqrt{1 + \frac{21}{4}}} = \frac{1}{2\sqrt{\frac{25}{4}}} = \frac{1}{2 \times \frac{5}{2}} = \frac{1}{5}$.

(C) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.
To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$.
Then $y+1 = (x+1)^2$,so $x+1 = \sqrt{y+1}$ (since $x \geq -1$),which gives $x = \sqrt{y+1} - 1$.
Thus,$f^{-1}(x) = \sqrt{x+1} - 1$.
We solve $f(x) = f^{-1}(x)$,which implies $(x+1)^2 - 1 = \sqrt{x+1} - 1$.
This simplifies to $(x+1)^2 = \sqrt{x+1}$.
Squaring both sides,we get $(x+1)^4 = x+1$,or $(x+1)((x+1)^3 - 1) = 0$.
This gives $x+1 = 0$ or $(x+1)^3 = 1$.
Case $1$: $x+1 = 0 \Rightarrow x = -1$.
Case $2$: $x+1 = 1 \Rightarrow x = 0$.
Case $3$: $x+1 = \omega$ or $x+1 = \omega^2$,where $\omega$ is a complex cube root of unity.
$x = \omega - 1 = \frac{-1+i\sqrt{3}}{2} - 1 = \frac{-3+i\sqrt{3}}{2}$ and $x = \omega^2 - 1 = \frac{-1-i\sqrt{3}}{2} - 1 = \frac{-3-i\sqrt{3}}{2}$.
Since the domain is $x \geq -1$,we only consider real values $x = -1$ and $x = 0$.
Thus,the set is $\{0, -1\}$.

(C) Given $f(x) = x - \frac{1}{x}$. Let $y = f(x)$,so $y = x - \frac{1}{x}$.
To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:
$y = \frac{x^2 - 1}{x} \Rightarrow x^2 - yx - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{y \pm \sqrt{y^2 - 4(1)(-1)}}{2(1)} = \frac{y \pm \sqrt{y^2 + 4}}{2}$.
Since the domain of $f$ is $[1, \infty)$,$x$ must be positive,so we take the positive root:
$x = \frac{y + \sqrt{y^2 + 4}}{2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x + \sqrt{x^2 + 4}}{2}$.

(C) Given $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$.
To find $f^{-1}(3)$,we set $f(x) = y$,which implies $x = f^{-1}(y)$.
$\frac{1+\sqrt{1+4 \log_2 x}}{2} = y$
$\sqrt{1+4 \log_2 x} = 2y - 1$
Squaring both sides:
$1 + 4 \log_2 x = (2y - 1)^2$
$1 + 4 \log_2 x = 4y^2 - 4y + 1$
$4 \log_2 x = 4y^2 - 4y$
$\log_2 x = y^2 - y$
$x = 2^{y^2 - y}$
Thus,$f^{-1}(y) = 2^{y^2 - y}$.
Now,substitute $y = 3$:
$f^{-1}(3) = 2^{3^2 - 3}$
$f^{-1}(3) = 2^{9 - 3}$
$f^{-1}(3) = 2^6 = 64$.

(A) Given the function $f(x) = 1 + 2x + 4x^2 + 8x^3 + \ldots$ for $|x| < \frac{1}{2}$.
This is an infinite geometric series with first term $a = 1$ and common ratio $r = 2x$.
The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.
Thus,$f(x) = \frac{1}{1-2x}$.
Let $y = f(x) = \frac{1}{1-2x}$.
To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:
$y(1-2x) = 1$
$y - 2xy = 1$
$2xy = y - 1$
$x = \frac{y-1}{2y}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x-1}{2x}$.

(A) Given $f(x) = 3x + \frac{2}{x}$. Let $y = 3x + \frac{2}{x}$.
Multiplying by $x$,we get $3x^2 - yx + 2 = 0$.
Using the quadratic formula,$x = \frac{y \pm \sqrt{y^2 - 4(3)(2)}}{2(3)} = \frac{y \pm \sqrt{y^2 - 24}}{6}$.
Since the domain is $x \in [1, \infty)$,we must choose the root that satisfies this condition.
For $y \geq 5$,$y^2 \geq 25$,so $\sqrt{y^2 - 24} \geq 1$.
If we take $x = \frac{y - \sqrt{y^2 - 24}}{6}$,then for $y=5$,$x = \frac{5 - 1}{6} = \frac{2}{3} < 1$,which is outside the domain.
If we take $x = \frac{y + \sqrt{y^2 - 24}}{6}$,then for $y=5$,$x = \frac{5 + 1}{6} = 1$,which is in the domain.
Thus,$f^{-1}(x) = \frac{x + \sqrt{x^2 - 24}}{6}$.

(A) We have a function $f: (-\infty, 0] \rightarrow [0, \infty)$ defined as $f(x) = x^2$.
Since each line parallel to the $x$-axis cuts the curve at most at one point,the function $f$ is one-one.
From the graph,it is clear that the range of $f$ is $[0, \infty)$,which is equal to the codomain.
Therefore,$f$ is onto.
Since $f$ is both one-one and onto,it is invertible.
The inverse function $f^{-1}$ maps the codomain of $f$ to the domain of $f$.
Thus,$f^{-1}: [0, \infty) \rightarrow (-\infty, 0]$.
Hence,the domain of $f^{-1}$ is $[0, \infty)$ and the range of $f^{-1}$ is $(-\infty, 0]$.

(D) Given that $g(x) = f^{-1}(x)$.
By the definition of inverse functions,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
Therefore,$f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$.
We are given $g(5) = 6$ and $g^{\prime}(5) = \frac{3}{4}$.
Substituting $x = 5$ into the equation $f^{\prime}(g(x)) = \frac{1}{g^{\prime}(x)}$,we get $f^{\prime}(g(5)) = \frac{1}{g^{\prime}(5)}$.
Since $g(5) = 6$,this becomes $f^{\prime}(6) = \frac{1}{g^{\prime}(5)}$.
Substituting the value $g^{\prime}(5) = \frac{3}{4}$,we get $f^{\prime}(6) = \frac{1}{3/4} = \frac{4}{3}$.

(C) Given,$y = \log_{a}(x + \sqrt{x^{2} + 1})$,$a > 0, a \neq 1$.
Taking exponential on both sides,we get $a^{y} = x + \sqrt{x^{2} + 1}$.
Now,consider $a^{-y} = \frac{1}{x + \sqrt{x^{2} + 1}}$.
Rationalizing the denominator,we get $a^{-y} = \sqrt{x^{2} + 1} - x$.
Subtracting the two equations: $a^{y} - a^{-y} = (x + \sqrt{x^{2} + 1}) - (\sqrt{x^{2} + 1} - x) = 2x$.
Thus,$x = \frac{a^{y} - a^{-y}}{2}$.
Since $a^{y} = e^{y \ln a}$,we have $x = \frac{e^{y \ln a} - e^{-y \ln a}}{2}$.
Using the definition $\sinh(u) = \frac{e^{u} - e^{-u}}{2}$,we get $x = \sinh(y \ln a)$.
Therefore,the inverse function is $f^{-1}(y) = \sinh(y \log a)$.

(A) Given that $f$ is the inverse function of $h$,we have $h(f(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$h^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$
Therefore,$f^{\prime}(x) = \frac{1}{h^{\prime}(f(x))}$.
Given $h^{\prime}(x) = \frac{1}{1 + \log x}$,we substitute $f(x)$ for $x$ to get $h^{\prime}(f(x)) = \frac{1}{1 + \log(f(x))}$.
Substituting this into the expression for $f^{\prime}(x)$:
$f^{\prime}(x) = \frac{1}{\frac{1}{1 + \log(f(x))}} = 1 + \log(f(x))$.

(A) Given that $f$ is the inverse function of $g$,we have $f(x) = g^{-1}(x)$.
We know that the derivative of an inverse function is given by the formula: $f^{\prime}(x) = \frac{1}{g^{\prime}(f(x))}$.
Given $g^{\prime}(x) = \frac{1}{1+x^n}$,we substitute $f(x)$ for $x$ in the expression for $g^{\prime}(x)$:
$g^{\prime}(f(x)) = \frac{1}{1+\{f(x)\}^n}$.
Now,substitute this into the formula for $f^{\prime}(x)$:
$f^{\prime}(x) = \frac{1}{\frac{1}{1+\{f(x)\}^n}} = 1+\{f(x)\}^n$.