The function $$ f $$ is defined as follows. $$ f(x)=\left\{\begin{array}{ll}3+x & ext { if } x

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The function $$ f $$ is defined as follows. $$ f(x)=\left\{\begin{array}{ll}3+x & 	ext { if } x<0 \ x^{2} & 	ext { if } x \geq 0\end{array}ight. $$ (a) Find the domain of the function. (b) Locate any intercepts. (c) Graph the function. (d) Based on the graph, find the range. (a) The domain of the function $$ f $$ is $$ (-\infty, \infty) $$. (Type your answer in interval notation.) (b) Locate any intercepts. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The intercept(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There are no intercepts.

UpStudy AI · Accepted Answer

**(a) Domain**

The function is defined by two pieces that cover all real numbers:  
$$
f(x)=
\begin{cases}
3+x & \text{if } x<0,\[5mm]
x^{2} & \text{if } x\geq 0.
\end{cases}
$$
Thus, the domain is  
$$
(-\infty, \infty).
$$

**(b) Intercepts**

1. **$$y$$-intercept:**  
   Set $$x=0$$. Since $$0\geq0$$, use the second branch:
   $$
   f(0)=0^2=0.
   $$
   So, the $$y$$-intercept is $$(0,0)$$.

2. **$$x$$-intercepts:**  
   To find the $$x$$-intercepts, solve $$f(x)=0$$.

- For $$x \geq 0$$ (using $$f(x)=x^2$$):
     $$
     x^2=0\quad \Longrightarrow \quad x=0.
     $$
     This gives the intercept $$(0,0)$$.

- For $$x<0$$ (using $$f(x)=3+x$$):
     $$
     3+x=0\quad \Longrightarrow \quad x=-3.
     $$
     This gives the intercept $$(-3,0)$$.

Thus, the intercepts are  
$$
(-3,0),\ (0,0).
$$

**(c) Graph of the Function**

- For $$x<0$$:  
  The function is a line given by 
  $$
  f(x)=3+x.
  $$
  This line has a slope of 1 and $$y$$-intercept $$3$$ (if extended beyond the defined interval). It passes through the point $$(-3,0)$$.

- For $$x\geq0$$:  
  The function is a parabola given by 
  $$
  f(x)=x^2.
  $$
  The parabola opens upward with its vertex at $$(0,0)$$.

On the graph, the line for $$x<0$$ will extend indefinitely to the left, while approaching a value of $$3$$ as $$x$$ approaches $$0$$ from the left (but never actually reaching $$x=0$$ in this branch). The parabola for $$x\geq0$$ begins at $$(0,0)$$ and continues upward.

A rough sketch would show:
- A line with a $$45^\circ$$ angle (slope 1) for $$x<0$$ passing through $$(-3,0)$$.
- A parabola starting at $$(0,0)$$ for $$x\geq0$$, curving upward.

**(d) Range**

- For the line ($$x<0$$):  
  As $$x\to -\infty$$, $$3+x\to -\infty$$; as $$x\to 0^{-}$$, $$3+x\to 3$$ (but does not include 3 since $$x=0$$ is not in this branch). Thus, the line part covers
  $$
  (-\infty,3).
  $$

- For the parabola ($$x\geq0$$):  
  Since $$f(x)=x^2$$ and $$x^2\geq0$$ for all $$x\geq0$$, the parabola covers
  $$
  [0,\infty).
  $$

Taking the union of the two parts, we have:
$$
(-\infty,3)\cup[0,\infty) = (-\infty,\infty)
$$
because the interval $$[0,3)$$ is common to both pieces.

Thus, the range of the function is  
$$
(-\infty, \infty).
$$
**(a) Domain:** $$ (-\infty, \infty) $$ **(b) Intercepts:** $$ (-3,0),\ (0,0) $$ **(c) Graph:** A line with a slope of 1 for $$x < 0$$ passing through $$(-3,0)$$ and a parabola $$f(x) = x^2$$ for $$x \geq 0$$ starting at $$(0,0)$$. **(d) Range:** $$ (-\infty, \infty) $$

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