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when x<-2, |x+2| and |x-2| both mods open up with a negative sign, so from negative infinity to 2, F(x) = -x-2 -x +1 = -2x

when, -2<x<2, only |x-2| opens with a negative sign, so F(x)= x+2 - x +2 = 4
when x>2, both parts open with positive signs, hence F(x) = x + 2 + x - 2 = 2x
lublana

Answer:

[tex]f(x)=\left\{\begin{matrix}2x &,if\ \ x >2 \\-2x &,if\ \ x < -2\\4&,if\ \ -2 \leq x \leq 2\end{matrix}\right.[/tex]

Range of f(x)=R+

Step-by-step explanation:

We are given that

[tex]f(x)=\mid x-2\mid +\mid x+2\mid [/tex]

We have to express given function in non modulus form , sketch the graph of f(x) and determine the range of f(x).

[tex]f(x)=\left\{\begin{matrix}x-2+x+2 &,if\ \ x >2 \\ -(x-2)-(x+2)&,if\ \ x < -2\\ x+2-x+2&,if\ \ -2 \leq x \leq 2\end{matrix}\right.[/tex]

[tex]f(x)=f(x)=\left\{\begin{matrix}2x &,if\ \ x >2 \\-2x &,if\ \ x < -2\\4&,if\ \ -2 \leq x \leq 2\end{matrix}\right.[/tex]

We can see that from given graph

The graph of function lies above the x - axis.

It means no negative values will  obtained.Hence , the range is set positive real numbers .

Range of f(x)=R+

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