Question: $f: x \mapsto 2 + \dfrac{3}{x^2 + 2x + 2}$, $x \in \mathbb{R}$.

(a) Show $f$ does not have an inverse.

Solution. For $2 < k < 5$, the horizontal line $y = k$ cuts the graph at more than one point, so $f$ is not 1-1. $\therefore f^{-1}$ does not exist.

(b) Find the smallest $k$ such that $f$ restricted to $x \geq k$ has an inverse.

Solution. The vertex of $x^2 + 2x + 2 = (x+1)^2 + 1$ is at $x = -1$. So the smallest $k = -1$.

(c) Find $f^{-1}(x)$ for the restricted domain $x \geq -1$.

Solution. Let $y = 2 + \dfrac{3}{(x+1)^2 + 1}$.

$\dfrac{3}{(x+1)^2 + 1} = y - 2$

$(x+1)^2 + 1 = \dfrac{3}{y - 2}$

$(x+1)^2 = \dfrac{3}{y - 2} - 1$

$x + 1 = \pm\sqrt{\dfrac{3}{y - 2} - 1}$. Since $x \geq -1$, take the positive root:

Tip: The $\pm$ resolution is the critical step. Use the domain restriction to decide.

Question: $f: x \mapsto \dfrac{x - 5}{x + 2}$, $x \neq -2$. Find $f^{-1}$ and state its domain.

Solution.

Let $y = \dfrac{x - 5}{x + 2}$.

Multiply both sides by $(x + 2)$: $y(x + 2) = x - 5$.

Expand: $yx + 2y = x - 5$.

Collect $x$-terms: $yx - x = -5 - 2y$, so $x(y - 1) = -5 - 2y$.

Divide: $x = \dfrac{-5 - 2y}{y - 1} = \dfrac{5 + 2y}{1 - y}$.

From the graph of $f$, $R_f = \mathbb{R} \setminus \{1\}$.

Range: $R_{f^{-1}} = D_f = \mathbb{R} \setminus \{-2\}$.

Tip: For a linear-fractional (Möbius) function $\dfrac{ax+b}{cx+d}$, the range is $\mathbb{R} \setminus \{a/c\}$ — the horizontal asymptote.

Question: $f: x \mapsto 2x^3$, $x \geq 1$. $g: x \mapsto 3x + 1$, $x \in \mathbb{R}$.

(a) Show $gf$ exists.

$R_f = [2, \infty)$ (from $f$ on $x \geq 1$, $f(1) = 2$, increasing).

$D_g = (-\infty, \infty)$.

Since $[2, \infty) \subseteq \mathbb{R}$, $gf$ exists. $\checkmark$

(b) Find $gf(x)$.

$gf(x) = g(f(x)) = g(2x^3) = 3(2x^3) + 1 = 6x^3 + 1$, for $x \geq 1$.

(c) Find the range of $gf$.

Chain: $D_f = [1, \infty) \xrightarrow{f} R_f = [2, \infty) \xrightarrow{g} R_{gf} = [7, \infty)$.

At $x = 2$ (the boundary of $R_f$): $g(2) = 7$. As $x \to \infty$, $g(x) \to \infty$. So $R_{gf} = [7, \infty)$.

Tip: Always chain $D_f \to R_f \to R_{gf}$ when computing composite ranges. Don't try to compute $R_{gf}$ directly from $gf(x)$ unless the algebra is trivial.

Question: Show that $f: x \mapsto \dfrac{x + a}{x - 1}$, $x \neq 1$, is self-inverse (where $a \neq 1$). Hence find $f^{99}(-4)$.

Step 1. Find $f^{-1}$:

Let $y = \dfrac{x + a}{x - 1}$.

$y(x - 1) = x + a \Rightarrow xy - y = x + a$.

$xy - x = y + a \Rightarrow x(y - 1) = y + a$.

$x = \dfrac{y + a}{y - 1}$.

So $f^{-1}(x) = \dfrac{x + a}{x - 1}$ — same expression and domain as $f$. $\therefore$ $f$ is self-inverse. $\checkmark$

Step 2. Compute $f^{99}(-4)$:

Since $f$ is self-inverse, $f^n(x) = x$ for even $n$ and $f^n(x) = f(x)$ for odd $n$.

$99$ is odd, so $f^{99}(-4) = f(-4) = \dfrac{-4 + a}{-4 - 1} = \dfrac{a - 4}{-5} = \dfrac{4 - a}{5}$.

Tip: Iteration cycles in self-inverse functions are the #1 way exam-setters test the concept. Watch for "$f^{1000}(\ldots)$" or "$f^{2025}(\ldots)$" questions.

Question: $f(x) = \begin{cases} x + 1, & 0 < x \leq 2 \\ 5 - x, & 2 < x \leq 4 \end{cases}$ and $f(x) = f(x + 4)$ for all real $x$.

(a) Find $f(1)$, $f(22)$, $f(15)$.

• $x = 1 \in (0, 2]$: $f(1) = 1 + 1 = 2$.
• $f(22) = f(22 - 5 \cdot 4) = f(2)$. $x = 2 \in (0, 2]$: $f(2) = 2 + 1 = 3$.
• $f(15) = f(15 - 3 \cdot 4) = f(3)$. $x = 3 \in (2, 4]$: $f(3) = 5 - 3 = 2$.

(b) Sketch $y = f(x)$ for $-4 \leq x \leq 8$.

• Draw $y = x + 1$ on $(0, 2]$ (line from $(0,1)$ to $(2,3)$).
• Draw $y = 5 - x$ on $(2, 4]$ (line from $(2,3)$ to $(4,1)$).
• This is one fundamental period — translate horizontally by $-4$, $+4$ to cover $[-4, 8]$.

Tip: For large $x$, subtract multiples of $T$ to bring $x$ into the fundamental period. Then apply the piecewise rule.

Question: Given $gf(x) = \dfrac{4\cos x - 1}{2\cos x - 1}$ with $R_{(gf)^{-1}} = \left[\dfrac{\pi}{2}, \pi\right]$, find $(gf)^{-1}\left(\dfrac{3}{2}\right)$.

Step 1. Let $t = (gf)^{-1}\left(\dfrac{3}{2}\right)$, so $gf(t) = \dfrac{3}{2}$.

Step 2. Cross multiply:

$2(4\cos t - 1) = 3(2\cos t - 1)$

$8\cos t - 2 = 6\cos t - 3$

$2\cos t = -1 \Rightarrow \cos t = -\dfrac{1}{2}$.

Step 3. Solve over the range:

$t = \dfrac{2\pi}{3}$ or $t = \dfrac{4\pi}{3}$.

Step 4. Apply the range constraint:

$R_{(gf)^{-1}} = \left[\dfrac{\pi}{2}, \pi\right]$. Only $t = \dfrac{2\pi}{3}$ is in this range.

Tip: Always use $R_{(gf)^{-1}} = D_{gf} = D_f$ to filter solutions when the equation has multiple roots.

Question: Describe a sequence of transformations that maps $y = \ln x$ to $y = \ln\left(\dfrac{2}{1 + x}\right)$.

Step 1. Simplify the target form using log laws:

Step 2. Build the chain from $y = \ln x$:

A. Translation 1 unit in the negative $x$-direction: $y = \ln x \to y = \ln(x + 1)$.

B. Reflection in the $x$-axis: $y = \ln(x + 1) \to y = -\ln(x + 1)$.

C. Translation $\ln 2$ units in the positive $y$-direction: $y = -\ln(x + 1) \to y = \ln 2 - \ln(x + 1) = \ln\left(\dfrac{2}{1 + x}\right)$.

Three transformations: A → B → C.

Tip: Always state each transformation with type, axis/direction, and magnitude. The order matters — swapping B and C in this example gives a different curve.