Piecewise Function 1Ad340

1. **State the problem:** We are given a piecewise function $$f(x)=\begin{cases} 2x, & x<0 \\ -x^2 - 4x + 1, & 02 \end{cases}$$ and need to determine which of the given statements about $f$ is true. 2. **Analyze the domain:** The function is defined for $x<0$, $02$. Notice that $x=0$ and $x=2$ are not included in the domain because the function pieces are strict inequalities. - Therefore, the domain is $(-\infty,0) \cup (0,2) \cup (2,\infty)$. - This means the domain is **not all real numbers** because $x=0$ and $x=2$ are excluded. 3. **Check if the function is increasing over its entire domain:** - For $x<0$, $f(x)=2x$ which is a linear function with slope $2>0$, so it is increasing on $(-\infty,0)$. - For $02$, $f(x)=\frac{1}{2}x + 3$, which is linear with positive slope $\frac{1}{2}$, so increasing on $(2,\infty)$. - Since $f$ decreases on $(0,2)$, it is **not increasing over its entire domain**. 4. **Check behavior as $x \to +\infty$:** - For $x>2$, $f(x) = \frac{1}{2}x + 3$. - As $x \to +\infty$, $f(x) \to +\infty$ because the linear term dominates. 5. **Check continuity:** - At $x=0$, the function is not defined, so it cannot be continuous there. - At $x=2$, the function is not defined, so it cannot be continuous there. - Therefore, $f$ is **not continuous** on its entire domain. **Summary of statements:** - "The function is increasing over its entire domain." — False. - "As $x$ approaches positive infinity, $f(x)$ approaches positive infinity." — True. - "The domain is all real numbers." — False. - "The function is continuous." — False. **Final answer:** The true statement is: **As $x$ approaches positive infinity, $f(x)$ approaches positive infinity.**