Question

Given a function f(x)=|x+1|-|x -2|

How many solutions has f(x) = a, in respect of a?
If a E {-3, 3}, number of solutions is
if a E (-3, 3) number of solutions is
if a E (-infinity, -3) U (3,infinity) number of solutions is

Answer 1

a=(x+1)-(x-2)

=x-x+1+2

a=3

Answer 2

Explanation:

`f(x)=/x+1/-/x-2/\\a=/x+1/-/x-2/\\\text{square both sides}\\a^(2)={(x+1)}^(2)+{(x-2)}^(2)-2(x+1)(x-2)\\a^(2)=x^(2)+2x+1+x^(2)-4x+4-2x^(2)+2x+4\\a^(2)=9\\a=√(9)\\a=+3\text{ or }-3`

`\text{if }a \in \{-3, 3\},\text{the number of solutions is 2}`

`\text{if }a \in \{-\infty, -3\}\bigcup\{\infty, 3\},\text{it is just like saying from infinity to infinity, which is still 2}`