Graph each piecewise-defined function.

Question

f (x) = {\begin{matrix} 2 x + 1 & if x \geq 0 \\ x & if x < 0 \end{matrix}

Accepted Answer

Hey, everyone here we are asked to graph the following function F of X is equal to four X minus two. If X is less than or equal to zero and two X, if X is greater than zero here we have four answer choice options. ABC and D, each of which display a graph representing our given function. So the key to solving this problem is to graph each interval of the domain separately. So we can start with the first part of our function where we have F of X is equal to four X minus two, which is defined for when X is less than or equal to zero. So we want to substitute in values of X that are within the domain main of this function. And so we can find ordered pairs. So we can start for when X is equal to zero. So substituting in 04 X, we have F of zero is equal to four times zero minus two. Simplifying we get negative two. And so we see one ordered pair is at zero and negative two. Now we can repeat this process for other values of X that are now less than zero. So we can choose when X is equal to negative one. So we have F of negative one, which gives four times negative one minus two. Simplifying we have negative six. And so we see a second ordered pair is at negative one and negative six repeating this process one more time 41, X is equal to negative two. We have F of negative two is equal to four times negative two minus two. Now, simplifying we get negative 10. So we see a third ordered pair is at negative two and negative 10. So now with our three different ordered pairs, the next step is to just graph these three points. So now we see we have our three points individually graphed. And so we have started graphing the first part of our original function. And so the next step is to just draw a line that connects each of these points. And so doing this, we have graphed our function F of X and we see that we have an end point at zero and negative two. And we see that this is a solid point since this, since X is defined for when X is equal to zero or the function is defined for when X is equal to zero. So again, we get a solid end point and then we see that the line extends to negative infinity because X is also less than zero. So now from here, we want to repeat this process. Now with the second interval from our original function where we now have F of X is equal to two X if X is greater than zero. So we want to start by having X B equal to zero. And thus substituting in 04 X in this function. So we have F of zero is equal to two times zero, which just gives us zero. So we see that one ordered pair for this second part of our function is at zero and zero. So now we want to repeat this process for values of X that are greater than zero. And so we can choose the value of one for X. So we have X is equal to one. And now finding F of one, we have two times one which just gives us two. So we see a second ordered pair is at one and two. And now repeating this process a third time, we can choose the value of two. So having X be equal to two, we find F of two is equal to two times two, which gives us four. So we see a third ordered pair is two and four. So now with our three new ordered pairs for the second part of our original function, we now want to repeat the similar process by graphing these points or plotting these points on a graph. And so we see that these three points are plotted on the same graph as the line that we found from earlier earlier, we now need to draw a line connecting these through these three new points. And so doing this, we now have graph. Now the second part of our original function and we see that we have an end point at 00 or at the origin, which is a hollow circle. And this is because we know that the function F of X is equal to two X is undefined for when X is equal to zero, since we want X to be greater than zero. So with this, we have graphed both parts of our function. So we have a finalized graph here, Sean where we again have one increasing line that starts at the origin and goes to positive infinity and has a hollow end point at the origin. And then our second line starts at negative two on the Y axis and goes all the way to negative infinity. So with this finalized graph, we see that we are left here with answer choice D. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

Graph each piecewise-defined function.
$f (x) = {\begin{matrix} 2 x + 1 & if x \geq 0 \\ x & if x < 0 \end{matrix}$

Key Concepts

Piecewise-Defined Functions

Graphing Linear Functions

Domain Restrictions and Continuity

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