Exercise 2 Let $$ f(x)=\left\{\begin{array}{ll}x^{2}+2 x, & ext { if } x \leq-1 \ x, & ext { if }-11\end{array} \quad ight. $$ a piecewise defined function. a) What is the domain? b) Find $$ f(-4), f\left(-\frac{3}{2} ight), f(-1), f(0), f(25) $$

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Exercise 2 Let $$ f(x)=\left\{\begin{array}{ll}x^{2}+2 x, & 	ext { if } x \leq-1 \ x, & 	ext { if }-11\end{array} \quadight. $$ a piecewise defined function. a) What is the domain? b) Find $$ f(-4), f\left(-\frac{3}{2}ight), f(-1), f(0), f(25) $$

UpStudy AI · Accepted Answer

**a) Domain**

The function is defined piecewise for all real numbers. Therefore, the domain is  
$$
\mathbb{R}.
$$

**b) Function Values**

1. For $$ x = -4 $$:  
Since $$-4 \leq -1$$, we use  
$$
f(x)=x^2+2x.
$$  
Thus,  
$$
f(-4)=(-4)^2+2(-4)=16-8=8.
$$

2. For $$ x = -\frac{3}{2} $$:  
Since $$-\frac{3}{2} \leq -1$$, we use  
$$
f(x)=x^2+2x.
$$  
Thus,  
$$
f\left(-\frac{3}{2}\right)=\left(-\frac{3}{2}\right)^2+2\left(-\frac{3}{2}\right)=\frac{9}{4}-3=\frac{9-12}{4}=-\frac{3}{4}.
$$

3. For $$ x = -1 $$:  
Since $$-1 \leq -1$$, we use  
$$
f(x)=x^2+2x.
$$  
Thus,  
$$
f(-1)=(-1)^2+2(-1)=1-2=-1.
$$

4. For $$ x = 0 $$:  
Since $$-1 < 0 \leq 1$$, we use  
$$
f(x)=x.
$$  
Thus,  
$$
f(0)=0.
$$

5. For $$ x = 25 $$:  
Since $$25 > 1$$, we use  
$$
f(x)=-1.
$$  
Thus,  
$$
f(25)=-1.
$$
**a) Domain** The domain of the function is all real numbers, $$\mathbb{R}$$. **b) Function Values** 1. $$ f(-4) = 8 $$ 2. $$ f\left(-\frac{3}{2}\right) = -\frac{3}{4} $$ 3. $$ f(-1) = -1 $$ 4. $$ f(0) = 0 $$ 5. $$ f(25) = -1 $$

Exercise 2 Let \( f(x)=\left\{\begin{array}{ll}x^{2}+2 x, & \text { if } x \leq-1 \\ x, & \text { if }-11\end{array} \quad\right. \) a piecewise defined function. a) What is the domain? b) Find \( f(-4), f\left(-\frac{3}{2}\right), f(-1), f(0), f(25) \)

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