1. **Stating the problem:** We have a function $f: \mathbb{R} \to \mathbb{Z}$ defined by $f(x) = \lfloor 2x - 1 \rfloor$, where $\lfloor \cdot \rfloor$ denotes the floor function (greatest integer less than or equal to the argument). We want to find $f(A)$ where $A = \{x \mid 1 \leq x \leq 4\}$. 2. **Understanding the function and domain:** The function takes any real number $x$ in the interval $[1,4]$, computes $2x - 1$, and then applies the floor function to get an integer. 3. **Calculate the range of $2x - 1$ over $A$:** - When $x=1$, $2(1) - 1 = 2 - 1 = 1$ - When $x=4$, $2(4) - 1 = 8 - 1 = 7$ So, $2x - 1$ ranges over $[1,7]$ as $x$ goes from 1 to 4. 4. **Apply the floor function to the interval $[1,7]$:** The floor function applied to values in $[1,7]$ will produce all integers from 1 up to 7 inclusive. 5. **Therefore, the image $f(A)$ is:** $$f(A) = \{1, 2, 3, 4, 5, 6, 7\}$$ This is because for each integer $k$ in $1$ to $7$, there exists some $x$ in $[1,4]$ such that $2x - 1$ is in $[k, k+1)$, so $f(x) = k$. **Final answer:** $$f(A) = \{1, 2, 3, 4, 5, 6, 7\}$$