Write the Function into piece wise defined function if f(x) = |x+1| + |x-2| + |x+3| +|x-9| ?

Question

anonymous · Answer

lol havfun
first see what you get if $x<-3$

primeralph · Answer

Just a bunch of combined inequalities.

anonymous · Answer

i got 5 piece wise defined function am i right ?

anonymous · Answer

yes, should be five

anonymous · Answer

Yup..Thxxx...::)

primeralph · Answer

@Yahoo!  You can try to condense them.

anonymous · Answer

What if the function is of this kind   f(x)= 2|x| + |x+2| - | | x+2| - 2|x| |

primeralph · Answer

It's all the same basic principle.

anonymous · Answer

But wat to do with   | | x+2| - 2|x| |

anonymous · Answer

Double Modulus

anonymous · Answer

@satellite73

anonymous · Answer

why you want to make me think?

anonymous · Answer

i tried.... but failed

anonymous · Answer

no that was wrong sorry

anonymous · Answer

if $x>0$ then $|x+2|=x+2$ and $2|x|=2x$ so 
$$|x+2-2x|=|-x+2|$$

anonymous · Answer

now if $x>2$ you have $|-x+2|=x-2$ so on the interval $2,\infty$ this is $x-2$ and on the interval $0,2$ this is $2-x$

anonymous · Answer

now if $x<-2$ you have $|x+2|=-x-2$  and $|x|=-x$ giving you 
$$|-x-2-2(-x)|=|x-2|$$

anonymous · Answer

and since $x<-2$ you know $|x-2|=2-x$

anonymous · Answer

there really must be an easier way to do this
you are not done
you have to consider what happens on the interval $(-2,0)$ but the work is still the same

anonymous · Answer

but one thing we know from above, it is 
$$|x-2|$$ on the inteval $(-\infty,-2)\cup (0\infty)$