Q.The function $f$ is defined by $f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}$ Draw the graph of $f(x)$.

Concept: Piecewise Function Graphing — check each piece's rule and then check whether the pieces agree at the boundary point before drawing anything.

• For $x < 0$: $f(x) = 1 - x$ — a line of slope $-1$, $y$-intercept $1$.
• At $x = 0$: $f(0) = 1$.
• For $x > 0$: $f(x) = x + 1$ — a line of slope $1$, $y$-intercept $1$. …

To graph a piecewise function, treat each rule separately over its own domain, then check what happens at the boundary point(s) before drawing the final picture.

Step 1 — Plot the piece for $x < 0$: $f(x) = 1 - x$.

This is a line of slope $-1$ and $y$-intercept $1$, drawn only for $x<0$.

• $x=-1 \Rightarrow f(-1) = 1-(-1) = 2$, so $(-1, 2)$ is on the graph.
• $x=-2 \Rightarrow f(-2) = 1-(-2) = 3$, so $(-2, 3)$ is on the graph.
• As $x \to 0^-$ (e.g. $x=-0.1$), $f(-0.1) = 1.1 \to 1$.

Step 2 — Plot the piece at $x = 0$: $f(0) = 1$.

This gives the single point $(0, 1)$.

Step 3 — Plot the piece for $x > 0$: $f(x) = x + 1$.

This is a line of slope $1$ and $y$-intercept $1$, drawn only for $x>0$.

• $x=1 \Rightarrow f(1) = 1+1 = 2$, so $(1, 2)$ is on the graph.
• $x=2 \Rightarrow f(2) = 2+1 = 3$, so $(2, 3)$ is on the graph.
• As $x \to 0^+$ (e.g. $x=0.1$), $f(0.1) = 1.1 \to 1$.

Step 4 — Check what happens at the join, $x = 0$.

The left piece approaches $1$ as $x \to 0^-$, the right piece approaches $1$ as $x \to 0^+$, and the middle piece gives $f(0) = 1$ directly. All three values are the same number, $1$. That means the two rays don't stop short of $(0,1)$ and leave a gap for the middle piece to fill — they run straight into it. The function is continuous at $x = 0$; no open circle is needed on either branch.

Step 5 — Recognise the closed form.

For $x < 0$: $|x| = -x$, so $|x| + 1 = -x + 1 = 1 - x$ — matches the first piece exactly.

At $x = 0$: $|0| + 1 = 1$ — matches the middle piece exactly.

For $x > 0$: $|x| = x$, so $|x| + 1 = x + 1$ — matches the third piece exactly.

So for every real $x$, $f(x) = |x| + 1$. The three-part definition is just a longer way of writing the familiar absolute-value graph shifted up by $1$ unit.

Step 6 — Draw the graph. …