Calculus, 8th Edition,stewart; page 21 problem 41

Evaluate f(-3), f(0), and f(2) for the piecewise defined function. Then sketch the graph of the function.

asked Jul 27, 2015 in CALCULUS by anonymous

Share this question

1 Answer

Step:1

The function is $f(x)=\left\{\begin{matrix} x+2 & \textup{\textrm{if}} & x< 0 \\ 1-x & \textrm{if} &x\geq 0 \end{matrix}\right.$ .

The function $f(x)$ is called piecewise defined function.

Find the domain of $f(x)=x+2$ .

If $x<0$ then the function is $f(x)=x+2$ .

All possible values of $x$ is the domain of the function.

The function does not have any radicals.

The domain of the function is the set of all real numbers.

Find the domain of $f(x)=1-x$ .

If $x\geq 0$ then the function is $f(x)=1-x$ .

All possible values of $x$ is the domain of the function.

The function does not have any radicals.

The domain of the function is the set of all real numbers.

Domain of the piecewise function $f(x)=\left\{\begin{matrix} x+2 & \textup{\textrm{if}} & x< 0 \\ 1-x & \textrm{if} &x\geq 0 \end{matrix}\right.$ is $(-\infty , \infty )$ .

Step:2

Graph of the function $f(x)=\left\{\begin{matrix} x+2 & \textup{\textrm{if}} & x< 0 \\ 1-x & \textrm{if} &x\geq 0 \end{matrix}\right.$ .

Hallow circle indicates that the point is excluded from the function.

Solid circle indicates that the point is included in the function.

Solution:

Domain of the piecewise function $f(x)=\left\{\begin{matrix} x+2 & \textup{\textrm{if}} & x< 0 \\ 1-x & \textrm{if} &x\geq 0 \end{matrix}\right.$ is $(-\infty , \infty )$ .

Graph of the function $f(x)=\left\{\begin{matrix} x+2 & \textup{\textrm{if}} & x< 0 \\ 1-x & \textrm{if} &x\geq 0 \end{matrix}\right.$ is

answered Jul 28, 2015 by dozey Mentor

Most popular tags

Calculus, 8th Edition,stewart; page 21 problem 41

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

Please log in or register to add a comment.

Related questions