Draw the grap of the function f(x)={1+x,-1≤x≤0\1-x,0

1.	Draw the grap of the function f(x)={1+x,-1≤x≤0\1-x,0
Answer» $\large\underline{\sf{Solution-}}$ $\begin{gathered}\begin{gathered}\bf \:Given \: f(x) = y = \begin{cases} &\sf{1 + x \: if \: - 1 \leqslant x \leqslant 0} \\ &\sf{1 - x \: if \: 0 < x \leqslant 1} \end{cases}\end{gathered}\end{gathered}$ Case :- 1 $\sf \: When \: y \: = \: x + 1 \: if \: - 1 \leqslant x \leqslant 0$ 1. Substituting 'x = 0' in the given equation, we get $\rm :\longmapsto\:y = 1 + 0$ $\bf\implies \:y = 1$ 2. Substituting 'x = - 0.5' in the given equation, we get $\rm :\longmapsto\:y = 1 - 0.5$ $\bf\implies \:y = 0.5$ 3. Substituting 'x = - 1' in the given equation, we get $\rm :\longmapsto\:y = 1 - 1$ $\bf\implies \:y = 0$ Hᴇɴᴄᴇ, ➢ Pair of points of the given equation are shown in the below table. $\begin{gathered}\boxed{\begin{array}{c\|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf - 0.5 & \sf 0.5 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}$ Now DRAW a GRAPH using the points (0 , 1), (- 0.5 , 0.5) & (- 1 , 0) ➢ See the attachment graph. (Red line) Case :- 2 $\sf \: When \: y = 1 - x \: if \: 0 < x \leqslant 1$ 1. Substituting 'x = 1' in the given equation, we get $\rm :\longmapsto\:y = 1 - 1$ $\bf\implies \:y = 0$ 2. Substituting 'x = 0.5' in the given equation, we get $\rm :\longmapsto\:y = 1 - 0.5$ $\bf\implies \:y = 0.5$ 3. Substituting 'x = 0.4' in the given equation, we get $\rm :\longmapsto\:y = 1 - 0.4$ $\bf\implies \:y = 0.6$ Hᴇɴᴄᴇ, ➢ Pair of points of the given equation are shown in the below table. $\begin{gathered}\boxed{\begin{array}{c\|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 0 \\ \\ \sf 0.5 & \sf 0.5 \\ \\ \sf 0.4 & \sf 0.6 \end{array}} \\ \end{gathered}$ Now draw a graph using the points (1 , 0), (0.5 , 0.5) & (0.4 , 0.6) ➢ See the attachment graph. (BLUE line)

Answer»

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

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

Case :- 1

$\sf \: When \: y \: = \: x + 1 \: if \: - 1 \leqslant x \leqslant 0$

1. Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:y = 1 + 0$

$\bf\implies \:y = 1$

2. Substituting 'x = - 0.5' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 0.5$

$\bf\implies \:y = 0.5$

3. Substituting 'x = - 1' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 1$

$\bf\implies \:y = 0$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf - 0.5 & \sf 0.5 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}$

Now DRAW a GRAPH using the points (0 , 1), (- 0.5 , 0.5) & (- 1 , 0)

➢ See the attachment graph. (Red line)

Case :- 2

$\sf \: When \: y = 1 - x \: if \: 0 < x \leqslant 1$

1. Substituting 'x = 1' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 1$

$\bf\implies \:y = 0$

2. Substituting 'x = 0.5' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 0.5$

$\bf\implies \:y = 0.5$

3. Substituting 'x = 0.4' in the given equation, we get

$\rm :\longmapsto\:y = 1 - 0.4$

$\bf\implies \:y = 0.6$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 0 \\ \\ \sf 0.5 & \sf 0.5 \\ \\ \sf 0.4 & \sf 0.6 \end{array}} \\ \end{gathered}$

Draw the grap of the function f(x)={1+x,-1≤x≤0\1-x,0

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