Suppose that the function $$ f $$ is defined, for all real numbers, as follows. $$ \begin{array}{ll}\frac{1}{2} x-2 & \text { if } x2\end{array} $$ Find $$ f(-1), f(0) $$, and $$ f(4) $$ $$ f(-1)=\square $$ $$ f(0)=\square $$

Question

Suppose that the function $$ f $$ is defined, for all real numbers, as follows. $$ \begin{array}{ll}\frac{1}{2} x-2 & \text { if } x<-1 \ (x+1)^{2}-2 & \text { if }-1 \leq x \leq 2 \ -1 & \text { if } x>2\end{array} $$ Find $$ f(-1), f(0) $$, and $$ f(4) $$ $$ f(-1)=\square $$ $$ f(0)=\square $$

UpStudy AI · Accepted Answer

We are given the function

$$
f(x)=
\begin{cases}
\frac{1}{2}x-2 & \text{if } x<-1,\[1mm]
(x+1)^2-2 & \text{if } -1\le x\le 2,\[1mm]
-1 & \text{if } x>2.
\end{cases}
$$

**Step 1. Evaluate $$ f(-1) $$:**

Since $$-1$$ satisfies $$-1\le x\le 2$$, we use

$$
f(-1)=( (-1)+1)^2-2.
$$

Calculate inside the parenthesis:

$$
-1+1 = 0.
$$

Then,

$$
(0)^2-2= 0-2=-2.
$$

Thus,

$$
f(-1)=-2.
$$

**Step 2. Evaluate $$ f(0) $$:**

Since $$0$$ is also in the interval $$[-1,2]$$, we use

$$
f(0)=(0+1)^2-2.
$$

Calculate inside the parenthesis:

$$
0+1 = 1.
$$

Then,

$$
(1)^2-2=1-2=-1.
$$

Thus,

$$
f(0)=-1.
$$

**Step 3. Evaluate $$ f(4) $$:**

Since $$4>2$$, we use the third piece:

$$
f(4)=-1.
$$

**Final Answers:**

$$
f(-1)=-2, \quad f(0)=-1, \quad f(4)=-1.
$$
$$ f(-1) = -2 $$, $$ f(0) = -1 $$, and $$ f(4) = -1 $$.

Suppose that the function \( f \) is defined, for all real numbers, as follows. \[ \begin{array}{ll}\frac{1}{2} x-2 & \text { if } x<-1 \\ (x+1)^{2}-2 & \text { if }-1 \leq x \leq 2 \\ -1 & \text { if } x>2\end{array} \] Find \( f(-1), f(0) \), and \( f(4) \) \( f(-1)=\square \) \( f(0)=\square \)

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