Functions (Advanced) | DSE - Wyatt's Notes

1. Question: Let $f(x) = \dfrac{x + 2}{x - 1}$ and $g(x) = 2x - 3$. Find $(f \circ g)(x)$ and its domain.

$(f \circ g)(x) = f(2x - 3) = \dfrac{2x - 3 + 2}{2x - 3 - 1} = \dfrac{2x - 1}{2x - 4}$.

$\mathrm{dom}(g) = \mathbb{R}$. $\mathrm{dom}(f) = \mathbb{R} \setminus \{1\}$ So $g(x) \neq 1$: $2x - 3 \neq 1 \implies x \neq 2$. Also $2x - 4 \neq 0 \implies x \neq 2$. $\mathrm{dom}(f \circ g) = \mathbb{R} \setminus \{2\}$.

1. Question: Find $f^{-1}$ for $f(x) = \dfrac{3x + 1}{x - 2}$ ($x \neq 2$).

Set $y = \dfrac{3x + 1}{x - 2}$. Then $y(x - 2) = 3x + 1 \implies yx - 2y = 3x + 1 \implies x(y - 3) = 2y + 1$.

f^{-1}(x) = \dfrac{2x + 1}{x - 3}$$\mathrm{dom}(f^{-1}) = \mathbb{R} \setminus \{3\}.

1. Question: Let $f(x) = x^2 - 4x + 3$ with domain $[1, \infty)$. Find $f^{-1}(0)$.

First find $f^{-1}$. Set $y = (x - 2)^2 - 1$. Since domain is $[1, \infty)$Range is $[-1, \infty)$.

$(x - 2)^2 = y + 1 \implies x - 2 = \sqrt{y + 1}$ (positive root since $x \geqslant 1$).

f^{-1}(x) = 2 + \sqrt{x + 1}$$\mathrm{dom}(f^{-1}) = [-1, \infty).

$f^{-1}(0) = 2 + \sqrt{0 + 1} = 2 + 1 = 3$.

Verification: $f(3) = 9 - 12 + 3 = 0$. Confirmed.

1. Question: The graph of $y = f(x)$ has a minimum at $(1, -2)$ and passes through $(0, 3)$. Find the corresponding points on $y = 3f(2x) + 1$.

$(1, -2) \to$ set 2x = 1 \implies x = 0.5$$y = 3(-2) + 1 = -5. Point: $(0.5, -5)$.

$(0, 3) \to$ set 2x = 0 \implies x = 0$$y = 3(3) + 1 = 10. Point: $(0, 10)$.

1. Question: A function $f$ is defined by $f(x) = 2 - x^2$ for $x \leqslant 0$. State the range of $f$ and find $f^{-1}$.

Since $x \leqslant 0$: $x^2 \geqslant 0$ So $f(x) = 2 - x^2 \leqslant 2$. As $x \to -\infty$ $f(x) \to -\infty$. Range: $(-\infty, 2]$.

Set $y = 2 - x^2 \implies x^2 = 2 - y \implies x = -\sqrt{2 - y}$ (negative root since $x \leqslant 0$).

f^{-1}(x) = -\sqrt{2 - x}$$\mathrm{dom}(f^{-1}) = (-\infty, 2].

1. Question: Given $f(x) = \sqrt{x - 1}$ and $g(x) = x^2 + x + 1$Find $\mathrm{dom}(g \circ f)$.

\mathrm{dom}(f) = [1, \infty)$$\mathrm{dom}(g) = \mathbb{R}.

$(g \circ f)(x) = g(\sqrt{x - 1}) = (\sqrt{x - 1})^2 + \sqrt{x - 1} + 1 = x - 1 + \sqrt{x - 1} + 1 = x + \sqrt{x - 1}$.

Since $g$ has no domain restriction, $\mathrm{dom}(g \circ f) = \mathrm{dom}(f) = [1, \infty)$.

1. Question: Find the domain and range of $f(x) = \dfrac{1}{x^2 + 1}$.

Domain: $x^2 + 1 \neq 0$ for all real $x$ (since $x^2 \geqslant 0$). $\mathrm{dom}(f) = \mathbb{R}$.

Range: $x^2 + 1 \geqslant 1$ So $0 \lt \dfrac{1}{x^2 + 1} \leqslant 1$. $\mathrm{range}(f) = (0, 1]$.

1. Question: Let $f(x) = |x - 3| + |x + 1|$. Express $f$ as a piecewise function and find its minimum value.

Critical points at $x = 3$ and $x = -1$:

$f(x) = \begin{cases} -(x - 3) + -(x + 1) = -2x + 2 & \mathrm{if } x \lt -1 \\ -(x - 3) + (x + 1) = 4 & \mathrm{if } -1 \leqslant x \leqslant 3 \\ (x - 3) + (x + 1) = 2x - 2 & \mathrm{if } x \gt 3 \end{cases}$

For $x \lt -1$: $f(x) = -2x + 2$Which is decreasing (as $x$ increases towards $-1$). As $x \to -1^-$: $f(x) \to 4$.

For $-1 \leqslant x \leqslant 3$: $f(x) = 4$ (constant).

For $x \gt 3$: $f(x) = 2x - 2$Which is increasing.

Minimum value: $4$Attained for all $x \in [-1, 3]$.

$\mathrm{dom}(g) = [3, \infty)$. $\mathrm{dom}(f) = \mathbb{R} \setminus \{2\}$.

For $f \circ g$: We need $g(x) \in \mathrm{dom}(f)$I.e., $\sqrt{x-3} \neq 2$.

$\sqrt{x-3} = 2 \implies x = 7$. So exclude $x = 7$.

$\mathrm{dom}(f \circ g) = [3, \infty) \setminus \{7\}$.

For $g \circ f$: We need $x \in \mathrm{dom}(f)$ and $f(x) \geq 3$.

$\frac{x+1}{x-2} \geq 3 \implies \frac{x+1-3(x-2)}{x-2} \geq 0 \implies \frac{-2x+7}{x-2} \geq 0$

Critical points: $x = \dfrac{7}{2}$ and $x = 2$.

At $x = \dfrac{7}{2}$: expression equals $0$Which satisfies $\geq 0$.

$\mathrm{dom}(g \circ f) = \left(2,\; \dfrac{7}{2}\right]$.

Complete the square: $f(x) = 2(x^2 - 4x) + 5 = 2\!\left[(x-2)^2 - 4\right] + 5 = 2(x-2)^2 - 3$.

Since the domain is $[2, \infty)$ and the vertex is at $x = 2$The function is strictly increasing and hence one-to-one.

Range: $[-3, \infty)$.

Set $y = 2(x-2)^2 - 3$:

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

$x = 2 + \sqrt{\frac{y+3}{2}}$

(positive root since $x \geq 2$)

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

\mathrm{dom}(f^{-1}) = [-3, \infty)$$\mathrm{range}(f^{-1}) = [2, \infty).

Verification: $f^{-1}(f(3)) = f^{-1}(-1) = 2 + \sqrt{1} = 3$. Correct.

Rewrite: $y = 2f(3(x-2)) + 5$.

For $(1, 4)$ on $y = f(x)$: Set $3(x-2) = 1$ So $x - 2 = \dfrac{1}{3}$Giving $x = \dfrac{7}{3}$.

$y = 2(4) + 5 = 13$

Corresponding point: $\left(\dfrac{7}{3},\; 13\right)$.

For the minimum at $(2, -1)$: Set $3(x-2) = 2$ So $x - 2 = \dfrac{2}{3}$Giving $x = \dfrac{8}{3}$.

$y = 2(-1) + 5 = 3$

Corresponding point: $\left(\dfrac{8}{3},\; 3\right)$. This is the minimum of the transformed graph.

$\mathrm{dom}(g) = \mathbb{R}$, $\mathrm{dom}(f) = (1, \infty)$.

$(f \circ g)(x) = f(g(x)) = \ln(x^2 + 1 - 1) = \ln(x^2)$.

Domain: need $g(x) \in \mathrm{dom}(f)$I.e., $x^2 + 1 \gt 1 \implies x^2 \gt 0 \implies x \neq 0$.

$\mathrm{dom}(f \circ g) = \mathbb{R} \setminus \{0\}$.

$(g \circ f)(x) = g(f(x)) = [\ln(x-1)]^2 + 1$.

Domain: $\mathrm{dom}(g \circ f) = \mathrm{dom}(f) = (1, \infty)$.

Set $y = \dfrac{3x - 2}{x - 3}$:

$y(x - 3) = 3x - 2 \implies xy - 3y = 3x - 2 \implies xy - 3x = 3y - 2$

$x(y - 3) = 3y - 2 \implies x = \frac{3y - 2}{y - 3}$

Interchanging $x$ and $y$:

$f^{-1}(x) = \frac{3x - 2}{x - 3} = f(x)$

Since $f^{-1} = f$The function is self-inverse.

$(f \circ g)(x) = f(\sqrt{x+2}) = \frac{2\sqrt{x+2} + 3}{\sqrt{x+2} - 1}$

$\mathrm{dom}(g) = [-2, \infty)$. $\mathrm{dom}(f) = \mathbb{R} \setminus \{1\}$.

Need $\sqrt{x+2} \neq 1 \implies x + 2 \neq 1 \implies x \neq -1$.

$\mathrm{dom}(f \circ g) = [-2, -1) \cup (-1, \infty)$.

Set $y = \dfrac{2x-1}{x+3}$: $y(x+3) = 2x-1 \implies xy + 3y = 2x - 1 \implies x(y-2) = -1 - 3y$.

$f^{-1}(x) = \frac{-1-3x}{x-2} = \frac{3x+1}{2-x}$

For $f(x) = f^{-1}(x)$:

$\frac{2x-1}{x+3} = \frac{3x+1}{2-x}$

$(2x-1)(2-x) = (3x+1)(x+3)$

$4x - 2x^2 - 2 + x = 3x^2 + 9x + x + 3$

$-2x^2 + 5x - 2 = 3x^2 + 10x + 3$

$-5x^2 - 5x - 5 = 0 \implies x^2 + x + 1 = 0$

$\Delta = 1 - 4 = -3 \lt 0$. No real solutions.

$f(x) = (x+2)^2 - 4$. Since $x \geq -2$ and the vertex is at $x = -2$, $f$ is strictly increasing.

Range: $[-4, \infty)$. Since $5 \geq -4$, $f^{-1}(5)$ exists.

Set $(x+2)^2 - 4 = 5 \implies (x+2)^2 = 9 \implies x + 2 = 3$ (positive root).

$x = 1$

$f^{-1}(5) = 1$. Verification: $f(1) = 1 + 4 = 5$. Correct.

$y = 2(3-x)^2 + 1 = 2[-(x-3)]^2 + 1 = 2(x-3)^2 + 1$.

1. Translate right by $3$ units: $y = (x-3)^2$.
2. Vertical stretch by factor $2$: $y = 2(x-3)^2$.
3. Translate up by $1$ unit: $y = 2(x-3)^2 + 1$.

The vertex moves from $(0, 0)$ to $(3, 1)$. The parabola opens upward in both cases.

First, $(g \circ f)(x) = g(f(x)) = \left(\dfrac{1}{x+1}\right)^2 = \dfrac{1}{(x+1)^2}$.

Then:

$(f \circ g \circ f)(x) = f\!\left(\frac{1}{(x+1)^2}\right) = \frac{1}{\dfrac{1}{(x+1)^2} + 1} = \frac{(x+1)^2}{(x+1)^2 + 1} = \frac{(x+1)^2}{x^2 + 2x + 2}$

Domain: need $x + 1 \neq 0 \implies x \neq -1$ And $\dfrac{1}{(x+1)^2} + 1 \neq 0$.

Since $\dfrac{1}{(x+1)^2} \geq 0$ for all $x \neq -1$The second expression is always at least $1 > 0$.

$\mathrm{dom}(f \circ g \circ f) = \mathbb{R} \setminus \{-1\}$.

Critical points: $2x - 1 = 0 \implies x = \dfrac{1}{2}$ And $x + 3 = 0 \implies x = -3$.

For $x \lt -3$: $f(x) = -(2x-1) + -(x+3) = -3x - 2$ (decreasing as $x$ increases).

For $-3 \leq x \lt \dfrac{1}{2}$: $f(x) = -(2x-1) + (x+3) = -x + 4$ (decreasing).

For $x \geq \dfrac{1}{2}$: $f(x) = (2x-1) + (x+3) = 3x + 2$ (increasing).

The minimum occurs at the transition from decreasing to increasing, i.e., at $x = \dfrac{1}{2}$:

$f\!\left(\frac{1}{2}\right) = 3\!\left(\frac{1}{2}\right) + 2 = \frac{7}{2}$

Minimum value: $\dfrac{7}{2}$Attained at $x = \dfrac{1}{2}$.

Let $y = \dfrac{x}{x^2 + 1}$. Then $yx^2 + y = x \implies yx^2 - x + y = 0$.

For real $x$This quadratic in $x$ must have $\Delta \geq 0$:

$\Delta = 1 - 4y^2 \geq 0 \implies y^2 \leq \frac{1}{4} \implies -\frac{1}{2} \leq y \leq \frac{1}{2}$

When $y = \dfrac{1}{2}$: $\dfrac{1}{2}x^2 - x + \dfrac{1}{2} = 0 \implies (x-1)^2 = 0 \implies x = 1$. Attainable.

When $y = -\dfrac{1}{2}$: $\dfrac{1}{2}x^2 + x + \dfrac{1}{2} = 0 \implies (x+1)^2 = 0 \implies x = -1$. Attainable.

Range: $\left[-\dfrac{1}{2},\; \dfrac{1}{2}\right]$.

$(g \circ f)(x) = g(2x - 1) = 2x - 1 + 3 = 2x + 2$.

$(f \circ h)(x) = f(h(x)) = 2h(x) - 1$.

Setting equal: $2h(x) - 1 = 2x + 2 \implies h(x) = x + \dfrac{3}{2}$.

Verification: $(f \circ h)(x) = 2\!\left(x + \dfrac{3}{2}\right) - 1 = 2x + 2$. Correct.