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

Given a function fxx1x 2 How many solutions has fx 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 3in class=

Respuesta :

Answer:

Step-by-step explanation:

[tex]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=\sqrt{9}\\a=+3\text{ or }-3[/tex]

[tex]\text{if }a \in \{-3, 3\},\text{the number of solutions is 2}[/tex]

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

Answer:

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

=x-x+1+2

a=3