Chapter T, Problem 6CDT

Let $f(x)=\{\begin{array}{c}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$

1. (a) Evaluate f(−2) and f(1).
2. (b) Sketch the graph of f.

To determine

To evaluate: The value of $f\left(-2\right)\text{and}f\left(1\right)$.

Answer to Problem 6CDT

The values of $f\left(-2\right)$ is $-3$ and $f\left(1\right)$ is 3.

Explanation of Solution

Given;

The functions is $f\left(x\right)=\{\begin{array}{l}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$.

Calculation:

Consider the given functions $f\left(x\right)=\{\begin{array}{l}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$.

The value of $x=-2$ is less then zero. So, the function is $f\left(x\right)=1-x^2$.

Substitute $x=-2$ in $f\left(x\right)=1-x^2$,

$\begin{array}{c}f\left(-2\right)=1-{\left(-2\right)}^2\\ =1-4\\ =-3\end{array}$

Thus, the value of $f\left(-2\right)$ is $-3$.

The value of $x=1$ is greater then zero. So, the function is $f\left(x\right)=2x+1$.

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

$\begin{array}{c}f\left(1\right)=2\times 1+1\\ =2+1\\ =3\end{array}$

Thus, the value of $f\left(1\right)$ is 3.

To determine

To sketch: The graph of f

Answer to Problem 6CDT

The value of $g\circ f$ is $2x^2+4x-5$.

Explanation of Solution

Given;

The functions is $f\left(x\right)=\{\begin{array}{l}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$.

Calculation:

The graph for the given function $f\left(x\right)=\{\begin{array}{l}1-x^2\\ 2x+1\end{array}$ is shown below in Figure 1.

From figure 1 it is observed that, the graph of the function $f\left(x\right)$ has an enclosed region on the negative x-axis.

The graph for the given function $f\left(x\right)=\{\begin{array}{l}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$ is shown below in Figure 2.

From Figure 2, it is observed that the graph of the function $f\left(x\right)=\{\begin{array}{l}1-x^2\text{if}x\le 0\\ 2x+1\text{if}x>0\end{array}$ is a continuous curve.