Question

find the domain and range of lx+1l

Answer 1

The modulus function $|x|$ is actually defined as,

$\longrightarrow |x|=\left\{\begin{array}{rr}x,&x\geq0\\-x,&x\leq0\end{array}\right.$

Replacing $x$ by $x+1$ we get,

$\longrightarrow |x+1|=\left\{\begin{array}{rr}x+1,&x+1\geq0\\-(x+1),&x+1\leq0\end{array}\right.$

$\longrightarrow |x+1|=\left\{\begin{array}{rr}x+1,&x\geq-1\\-x-1,&x\leq-1\end{array}\right.$

However, the domain of this function is,

$\longrightarrow x\in(-\infty,\ -1]\cup[-1,\ \infty)$

$\longrightarrow\underline{\underline{x\in\mathbb{R}}}$

For $x\geq-1,$

$\longrightarrow x\in[-1,\ \infty)$

Adding 1,

$\longrightarrow x+1\in[-1+1,\ \infty)$

Since $|x+1|=x+1$ for $x\geq-1,$

$\longrightarrow |x+1|\in[0,\ \infty)\quad\quad\dots(1)$

For $x\leq-1,$

$\longrightarrow x\in(-\infty,\ -1]$

Multiplying by -1, (note the limit change in interval)

$\longrightarrow -x\in[1,\ \infty)$

Subtracting 1,

$\longrightarrow -x-1\in[1-1,\ \infty)$

Since $|x+1|=-x-1$ for $x\leq-1,$

$\longrightarrow |x+1|\in[0,\ \infty)\quad\quad\dots(2)$

From (1) and (2) we get,

$\longrightarrow\underline{\underline{|x+1|\in[0,\ \infty)}}$

This is the range of the function.

Answer 2

Answer:

answer is [0,∞]

Step-by-step explanation:

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