The function $$ f $$ is defined as follows. $$ f(x)=\left\{\begin{array}{ll}4+2 x & ext { if } x

Question

The function $$ f $$ is defined as follows. $$ f(x)=\left\{\begin{array}{ll}4+2 x & 	ext { if } x<0 \ x^{2} & 	ext { if } x \geq 0\end{array}ight. $$ (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 for all real numbers. Thus, the domain is  
$$
(-\infty,\infty).
$$

**(b) Intercepts**

1. **x-intercepts:**  
   Solve $$ f(x)=0 $$.

- For $$ x<0 $$, $$ f(x)=4+2x $$. Set
     $$
     4+2x=0 \quad\Longrightarrow\quad 2x=-4 \quad\Longrightarrow\quad x=-2.
     $$
     Since $$ -2<0 $$, one x-intercept is  
     $$
     (-2,0).
     $$
   - For $$ x\ge0 $$, $$ f(x)=x^2 $$. Set
     $$
     x^2=0 \quad\Longrightarrow\quad x=0.
     $$
     This gives another intercept at  
     $$
     (0,0).
     $$

2. **y-intercept:**  
   Evaluate $$ f(0) $$. Here, since $$ 0\ge0 $$,
   $$
   f(0)=0^2=0,
   $$
   so the y-intercept is  
   $$
   (0,0).
   $$

Thus, the intercepts are $$(-2,0)$$ and $$(0,0)$$.

**(c) Graphing the Function**

1. **For $$ x<0 $$:**  
   The function is $$ f(x)=4+2x $$, a line with slope $$2$$ and y-intercept $$4$$.  
   - As $$ x\to0^- $$, $$ f(x)\to4 $$ but note that $$ x=0 $$ is not included in this piece. An open circle is placed at $$ (0,4) $$.  
   - For example, when $$ x=-2 $$, $$ f(-2)=4+2(-2)=0 $$; hence the point $$ (-2,0) $$ is on the graph.

2. **For $$ x\ge0 $$:**  
   The function is $$ f(x)=x^2 $$, a parabola opening upward with vertex at $$ (0,0) $$.  
   - The point $$ (0,0) $$ is included in this piece (filled point).

When sketching the graph, represent the line $$ y=4+2x $$ for $$ x<0 $$ with an open circle at $$ (0,4) $$ and the parabola $$ y=x^2 $$ for $$ x\ge0 $$ starting at $$ (0,0) $$.

**(d) Range**

1. **For $$ x<0 $$:**  
   The linear piece $$ f(x)=4+2x $$ is defined for $$ x<0 $$. As $$ x\to-\infty $$,
   $$
   f(x)=4+2x\to-\infty,
   $$
   and as $$ x\to0^- $$,
   $$
   f(x)\to4 \quad (\text{but } f(x)\neq4).
   $$
   Thus, the range for this piece is  
   $$
   (-\infty,4).
   $$

2. **For $$ x\ge0 $$:**  
   The quadratic piece $$ f(x)=x^2 $$ yields values
   $$
   f(x)\ge0,
   $$
   so its range is  
   $$
   [0,\infty).
   $$

Taking the union of the two pieces, the overall range is  
$$
(-\infty,4)\cup[0,\infty)=(-\infty, \infty).
$$
**(a) Domain:** All real numbers $$ (-\infty,\infty) $$ **(b) Intercepts:** - x-intercepts: $$(-2,0)$$ and $$(0,0)$$ - y-intercept: $$(0,0)$$ **(c) Graph:** - For $$ x<0 $$: A line $$ y=4+2x $$ with an open circle at $$ (0,4) $$. - For $$ x\ge0 $$: A parabola $$ y=x^2 $$ starting at $$ (0,0) $$. **(d) Range:** All real numbers $$ (-\infty,\infty) $$

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