Question

Draw the graph of the function f(x) = |x| + |x-1|

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Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = |x| + |x - 1|$

Let first define the function.

Here critical points are x = 0 and x = 1

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: if \: x \: < \: 0} \\ &\sf{ \: \: \: x \: \: if \: x \: \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition, f(x) is defined as

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - x - (x - 1) \: \: when \: x \: < \: 0} \\ \\ &\sf{x - (x - 1) \: when \: 0 \leqslant x < 1}\\ \\ &\sf{x + (x - 1) \: \: when \: x \geqslant 1} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - 2x + 1 \: \: when \: x \: < \: 0} \\ \\ &\sf{ \: \: 1 \: \: \: \: \: \: when \: 0 \leqslant x < 1}\\ \\ &\sf{2x- 1 \: \: when \: x \geqslant 1} \end{cases}\end{gathered}\end{gathered}$

So, we have three cases now.

Case :- 1

$\red{\rm :\longmapsto\:y = f(x) = - 2x + 1 \: \: when \: x < 0}$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf 3 \\ \\ \sf - 2 & \sf 5 \\ \\ \sf - 3 & \sf 7 \end{array}} \\ \end{gathered}$

Case :- 2

$\red{\rm :\longmapsto\:y = f(x) = 1 \: \: when \: 0 \leqslant x < 1}$

Its a line parallel to x - axis passes through (0, 1)

Case :- 3

$\red{\rm :\longmapsto\:y = f(x) = 2x - 1 \: \: when \: x \geqslant 1}$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 1 \\ \\ \sf 2 & \sf 3 \\ \\ \sf 3 & \sf 5 \end{array}} \\ \end{gathered}$