关于函数$f: \mathbb{R} \to \mathbb{Z}$（$f(x)=\lceil 2x-1 \rceil$）的系列问题咨询

Alright, let's work through each part of this problem step by step—you mentioned confusion about injectivity and inverse functions for this ceiling function, so we'll make sure to clarify those concepts as we go.

i. Is $f$ injective? #

First, let's recall what an injective (one-to-one) function is: a function is injective if no two distinct inputs give the same output. In other words, if $f(a) = f(b)$, then $a$ must equal $b$.

For our function $f(x) = \lceil 2x - 1 \rceil$, it's easy to find counterexamples. Let's pick $x=0$ and $x=0.2$:

• $f(0) = \lceil 2(0) - 1 \rceil = \lceil -1 \rceil = -1$
• $f(0.2) = \lceil 2(0.2) - 1 \rceil = \lceil -0.6 \rceil = -1$

Here, $0 \neq 0.2$, but $f(0) = f(0.2)$. That violates the injective rule, so $f$ is not injective.

ii. Find $f(A)$ where $A = {x \mid 1 \leq x \leq 4}$ #

First, let's find the range of the expression inside the ceiling function: $2x - 1$. When $x$ is between 1 and 4:

• At $x=1$, $2x-1 = 2(1)-1 = 1$
• At $x=4$, $2x-1 = 2(4)-1 =7$

So $2x-1$ covers every real number from 1 up to 7. The ceiling function $\lceil y \rceil$ takes every integer value starting from $\lceil 1 \rceil=1$ up to $\lceil7\rceil=7$, and hits every integer in between (for any integer $k$ between 1 and 7, there's an $x$ such that $2x-1$ falls in $(k-1, k]$, making $\lceil 2x-1 \rceil=k$).

Thus, $f(A) = {1,2,3,4,5,6,7}$.

iii. Find $f(B)$ where $B = {3,4,5,6,7}$ #

We just need to calculate $f(x)$ for each element in $B$ directly:

• $f(3) = \lceil 2(3) -1 \rceil = \lceil5\rceil=5$
• $f(4) = \lceil2(4) -1\rceil=\lceil7\rceil=7$
• $f(5) = \lceil2(5) -1\rceil=\lceil9\rceil=9$
• $f(6) = \lceil2(6) -1\rceil=\lceil11\rceil=11$
• $f(7) = \lceil2(7) -1\rceil=\lceil13\rceil=13$

So $f(B) = {5,7,9,11,13}$.

iv. Find $f^{-1}(C)$ where $C = {-9,-8}$ #

The inverse image $f^{-1}(C)$ is the set of all real numbers $x$ such that $f(x)$ is in $C$. Remember a key property of the ceiling function: for any integer $k$, $\lceil y \rceil = k$ if and only if $k-1 < y \leq k$. We'll apply this to $y=2x-1$ for each value in $C$:

1. For $k=-9$:
   $-9-1 < 2x-1 \leq -9$
   $-10 < 2x-1 \leq -9$
   Add 1 to all parts: $-9 < 2x \leq -8$
   Divide by 2: $-4.5 < x \leq -4$

2. For $k=-8$:
   $-8-1 < 2x-1 \leq -8$
   $-9 < 2x-1 \leq -8$
   Add 1: $-8 < 2x \leq -7$
   Divide by 2: $-4 < x \leq -3.5$

Combine these two intervals (they meet at $x=-4$, which is included in the first interval), so $f^{-1}(C) = (-4.5, -3.5]$.

v. Find $f^{-1}(D)$ where $D = {0.4,0.5,0.6}$ #

The function $f$ maps real numbers to integers (since $f: \mathbb{R} \to \mathbb{Z}$). The ceiling function $\lceil 2x-1 \rceil$ will always output an integer—none of the elements in $D$ are integers. That means there are no real numbers $x$ such that $f(x)$ equals any element of $D$.

Thus, $f^{-1}(D) = \emptyset$ (the empty set).

内容的提问来源于stack exchange，提问作者violet