函数 f(x) 的图像经历了以下变换： 步骤 1：沿 x 轴负方向平移 1 个单位 步骤 2：沿 y 轴正方向平移 5 个单位 步骤 3：以比例因子 $\tfrac{1}{2}$ 平行于 x 轴进行伸缩 所得图像的函数为 $h(x)=4x^{2}+4x+9$。用 $f(x)$ 表示 $h(x)$。由此，求出函数 $f(x)$ 的表达式。

1. Express $h(x)$ in terms of $f(x)$ by applying the transformations sequentially.

Let the initial function be $y = f(x)$.

Step 1: Translation in the negative $x$-direction by $1$ unit (replace $x$ with $x + 1$):
$y = f(x + 1)$

Step 2: Translation in the positive $y$-direction by $5$ units (add $5$ to the function):
$y = f(x + 1) + 5$

Step 3: Scaling parallel to the $x$-axis by a factor of $\frac{1}{2}$ (replace $x$ with $2x$):
$h(x) = f(2x + 1) + 5$

2. Equate the derived expression for $h(x)$ to the given quadratic equation and isolate the $f$ term.

Given $h(x) = 4x^2 + 4x + 9$, we equate the two expressions:
$f(2x + 1) + 5 = 4x^2 + 4x + 9$
$f(2x + 1) = 4x^2 + 4x + 4$

3. Find an expression for $f(x)$ by rewriting the right-hand side in terms of $(2x + 1)$.

Notice that $4x^2 + 4x + 4$ contains a perfect square. We can rewrite it in terms of $(2x + 1)$:
$f(2x + 1) = (4x^2 + 4x + 1) + 3$
$f(2x + 1) = (2x + 1)^2 + 3$

Replacing the expression $(2x + 1)$ with $x$, we obtain the original function:
$f(x) = x^2 + 3$

1. Evaluate the individual ceiling functions

$\lceil 2.5 \rceil = 3$
$\lceil 3.09 \rceil = 4$

2. Substitute the evaluated values into the expression and calculate the total

$2\lceil 2.5 \rceil + \lceil 3.09 \rceil = 2(3) + 4 = 6 + 4 = 10$ (shown)

1. Evaluate $g(0.12)$ by substituting $x = 0.12$ into the function

$g(0.12) = \left\lceil \frac{0.12 + 1}{2} \right\rceil = \lceil 0.56 \rceil = 1$

2. Evaluate $g(3)$ by substituting $x = 3$ into the function

$g(3) = \left\lceil \frac{3 + 1}{2} \right\rceil = \lceil 2 \rceil = 2$

3. Calculate the sum of $g(0.12)$ and $g(3)$

$g(0.12) + g(3) = 1 + 2 = 3$

1. Determine the intervals of $x$ for which $g(x)$ takes integer values within the given domain $0 < x \le 4$.

To find where $g(x) = 1$:
$0 < \frac{x+1}{2} \le 1 \implies 0 < x+1 \le 2 \implies -1 < x \le 1$
Restricting to the given domain $0 < x \le 4$, $g(x) = 1$ for $0 < x \le 1$.

To find where $g(x) = 2$:
$1 < \frac{x+1}{2} \le 2 \implies 2 < x+1 \le 4 \implies 1 < x \le 3$

To find where $g(x) = 3$:
$2 < \frac{x+1}{2} \le 3 \implies 4 < x+1 \le 6 \implies 3 < x \le 5$
Restricting to the given domain $0 < x \le 4$, $g(x) = 3$ for $3 < x \le 4$.

2. Describe the features of the graph to be sketched.

Based on the intervals, sketch three horizontal line segments:

• A segment at $y=1$ for $0 < x \le 1$, with an open circle at $(0, 1)$ and a solid dot at $(1, 1)$.
• A segment at $y=2$ for $1 < x \le 3$, with an open circle at $(1, 2)$ and a solid dot at $(3, 2)$.
• A segment at $y=3$ for $3 < x \le 4$, with an open circle at $(3, 3)$ and a solid dot at $(4, 3)$.

1. Determine the range of values for the expression inside the ceiling function.

Given $x > 0$, we have $x + 1 > 1$, which means $\frac{x+1}{2} > 0.5$.
The expression inside the ceiling function can take any real value greater than $0.5$.

2. Apply the ceiling function to find the range.

The ceiling function $\lceil y \rceil$ rounds $y$ up to the nearest integer.
For any real number $y > 0.5$, the possible values of $\lceil y \rceil$ are the integers $1, 2, 3, \dots$
Therefore, the range of $g(x)$ is the set of all positive integers.