Build the graph of a function with absolute value.

Question

The function is:

And my idea of graphic (i did it using two graphs and deleting some parts)

Is it correct?

Answer 1

The green parts are correct, but try to look at the blue part again. For example, we have $$ f(1) = |1-1| + |1+2| = 3. $$

Answer 2

Your choice of scale makes the graph a bit difficult to interpret, but it appears you correctly graphed $y = x^2 + 3$ when $|x| > 1$. However, you incorrectly graphed $y = |x + 2| + |x - 1|$ in the interval $[-1, 1]$.

Since $x - 1 \geq 0$ if $x \geq 1$ and $x - 1 < 0$ if $x < 1$, \begin{align*} |x - 1| & = \begin{cases} x - 1 & \text{if $x \geq 1$}\\ -(x - 1) & \text{if $x < 1$} \end{cases}\\ & = \begin{cases} x - 1 & \text{if $x \geq 1$}\\ -x + 1 & \text{if $x < 1$} \end{cases} \end{align*} Since $x + 2 \geq 0$ if $x \geq -2$ and $x + 2 < 0$ if $x < -2$, \begin{align*} |x + 2| & = \begin{cases} x + 2 & \text{if $x \geq -2$}\\ -(x + 2) & \text{if $x < -2$} \end{cases}\\ & = \begin{cases} x - 1 & \text{if $x \geq -2$}\\ -x - 2 & \text{if $x < -2$} \end{cases} \end{align*} Putting these rules together yields \begin{align*} |x + 2| + |x - 1| & = \begin{cases} x + 2 + x - 1 & \text{if $x \geq 1$}\\ x + 2 - (x - 1) & \text{if $-2 \leq x < 1$}\\ -(x + 2) - (x - 1) & \text{if $x < -2$} \end{cases}\\ & = \begin{cases} 2x + 1 & \text{if $x \geq 1$}\\ 3 & \text{if $-2 \leq x < 1$}\\ -2x - 1 & \text{if $x < 1$} \end{cases} \end{align*} If $|x| \leq 1$, then $-1 \leq x \leq 1$. If $-1 \leq x < 1$, then the rule for $-2 \leq x < 1$ applies, so $$y = |x + 2| + |x - 1| = 3$$ If $x = 1$, then the rule for $x \geq 1$ applies, so $$y = |x + 2| + |x - 1| = 2x + 1 = 2 \cdot 1 + 1 = 2 + 1 = 3$$ Thus, in the interval $[-1, 1]$, the graph of $y = |x + 2| + |x - 1|$ lies on the horizontal line $y = 3$.

The graph of the function $$ y = \begin{cases} x^2 + 3 & \text{if $|x| > 1$}\\ |x + 2| + |x - 1| & \text{if $|x| \leq 1$} \end{cases} $$ is sketched below. While the grid lines make it somewhat difficult to tell, the circles at the points $(1, 3)$ and $(-1, 3)$ are filled since these points are on the graph while the circles at the points $(1, 4)$ and $(1, -4)$ are empty since these points are not on the graph.