Question

Draw the graph of function f(x) = |x+1|. Check injectivity and surjectivity . Write its domain and range?​

Answer 1

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

Given function is

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

We know,

Definition of Modulus function is as

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

Thus, using this definition, we have

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

further can be rewritten as

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x + 1| = \begin{cases} &\sf{ - (x + 1)\: \: when \: x < - 1} \\ \\ &\sf{ \: \: x + 1 \: \: when \: x \geqslant - 1} \end{cases}\end{gathered}\end{gathered}$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 2 \\ \\ \sf - 1 & \sf 0\\ \\ \sf - 2 & \sf 1\\ \\ \sf 3 & \sf 2 \end{array}} \\ \end{gathered}$

[ See the attachment for the graph ].

Now,

We know,

One - One or injective function :- A function f(x) is said to be one one iff corresponding to one value of x, there is one value of f(x)

OR

$\sf \: f : a \to \: b \: is \: one - one \: if \: f(x) = f(y)\rm\implies \:x = y \: \forall \: x,y \in \: a$

So, here for the function f(x) = |x + 1|, we have

$\rm :\longmapsto\:f( - 2) = f(0) = 1$

$\bf\implies \:f(x) \: is \: not \: injective$

Now,

Onto or Surjective function :-

$\sf \: f : a \to \: b \: is \: onto \: if \forall \:y \in \: b \: there \: exist \: x \in \: a \: such \: that \: f(x) = y$

Now, here for the function f(x) = |x + 1|,

$\rm :\longmapsto\:f \: is \: not \: surjective \: as \: for \: - 1 \in \: real \: number, \\ \rm \: there \: doesnot \: exist \: any \: x \in \: real \: number.$

Hence, f(x) = | x + 1 | is neither injective nor Surjective function.