Hàm số, nhánh nghịch đảo và hợp hàm
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| injective/ɪnˈdʒektɪv/ | One-to-one: different inputs always give different outputs. |
| surjective/sɜːˈdʒektɪv/ | Onto: every element of the codomain is reached. |
| involution/ɪnvəˈluːʃn/ | phép tự đảo |
One output, two inputs
- $f(x) = (x-2)^2 - 5$ sends both $-1$ and $5$ to $4$. Asked for $f^{-1}(4)$, you need to know which input is allowed.
- GRE function questions test injective 单射, surjective 满射, inverse branches, involutions 对合 and iterated composition — always against the stated domain and codomain.
Injective, surjective, bijective
- Injective: equal outputs force equal inputs. Surjective: every element of the codomain is reached.
- $x \mapsto x^2$: from $\mathbb R$ to $[0,\infty)$ surjective, not injective; from $[0,\infty)$ to $[0,\infty)$ bijective; from $[0,\infty)$ to $\mathbb R$ injective, not surjective.
What kind of map?
Test injectivity and surjectivity against the stated sets.
Which map is a bijection?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
e^x misses negatives; x² fails injectivity or surjectivity.
Inverse branches
- Solve $y = f(x)$ for $x$, then let the original domain choose the branch.
$f(x) = (x-2)^2 - 5$ on $x \ge 2$: $x = 2 \pm\sqrt{y+5}$; the domain keeps $+$, so $f^{-1}(y) = 2 + \sqrt{y + 5}$ for $y \ge -5$. Then $f^{-1}(4) = 5$. Check: $f(5) = 4$.
$f(x) = x^2 + 4x$ on $x \ge -2$. Find $f^{-1}(5)$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$x^2 + 4x - 5 = 0$ gives 1 or −5; only 1 is allowed.
Reflection and involutions
- The inverse graph is the reflection in $y = x$ — not the reciprocal $1/f$.
- An involution satisfies $f(f(x)) = x$: $-x$, $1/x$, $c - x$. A strictly increasing involution on an interval must be the identity.
The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. Each point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$. Replacing $y$ by $1/y$ gives the reciprocal, a different function.
A function with $f(f(x)) = x$ for all $x$ in its domain is called an ____.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
It is its own inverse.
Iterated composition
- $f^n$ means $f \circ \cdots \circ f$, not $(f(x))^n$.
- $g(x) = \frac{1}{1-x}$: $g^2(x) = \frac{x-1}{x}$, $g^3(x) = x$. So $3 \to -\tfrac12 \to \tfrac23 \to 3$, and $g^8(3) = g^2(3) = \tfrac23$ (reduce $8$ mod 3).
A formula that cancels to $x$ does not restore excluded inputs: $g^3(x) = x$ still needs $x \neq 0, 1$.
$g(x) = \frac{1}{1-x}$. Find $g^{10}(2)$, the tenth iterate.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Period 3, 10 ≡ 1: $g(2) = -1$.
Recap
- Surjectivity depends on the stated codomain.
- Use the domain to choose the inverse branch; check both compositions.
- Find the period of an iteration and reduce the count.
- injective: One-to-one: different inputs always give different outputs.
- surjective: Onto: every element of the codomain is reached.