Q. 25 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=\left\{\begin{array}{l}{x}^2 if x<0\\ 2 if x=0\\ 2x+1 if x>0\end{array}\right.$

find,

$$\begin{gathered} a) f(-2\left) \\ b\right) f\left(0\right) \\ c) f(2) \end{gathered}$$

Step-by-Step Solution

Answer

Part (a) the value of $f(-2)$ is $4$. 

Part (b) the value of $f\left(0\right)$ is $2$.

Part (c) the value of  $f\left(2\right)$ is $5$.

Step 1. Given information.

The given function is,

$f\left(x\right)=\left\{\begin{array}{l}{x}^2 if x<0\\ 2 if x=0\\ 2x+1 if x>0\end{array}\right.$ 

Step 2. Solving Part (a)

We observe here $x=-2$that is $x<0$, then the function will be

$f\left(x\right)=x^2$

On substituting $x=-2$ we get,

$$\begin{gathered} f(-2)={\left(-2\right)}^2 \\ =4 \end{gathered}$$

Therefore the value,

$f(-2)=4$.

Step 3. Solving Part (b).

We observe here $x=0$, then the function will be,

$f\left(x\right)=2$

On substituting $x=0$, we get

$f\left(0\right)=2$

Therefore, the value

$f\left(0\right)=2$.

Step 4. Solving Part (c).

We observe here $x=2$ that is $x>0$, then the function will be

$f\left(x\right)=2x+1$

On substituting $x=2$, we get

$$\begin{gathered} f\left(2\right)=2\left(2\right)+1 \\ =4+1 \\ =5 \end{gathered}$$

Therefore, the value

$f\left(2\right)=5$.