## anonymous 3 years ago h(3) h(x)= 5|x|/x

1. anonymous

@Aylin

2. anonymous

i plug this one in calculator and got 5 as final answer

3. anonymous

That looks right.

4. anonymous

alright cool :D

5. anonymous

for the next one it's asking for h(-2/3) but you can't use a negative for absolute value right? or is it ?

6. anonymous

umm why over 5 ? isn't it over -2/3 ?

7. anonymous

@Shido88 Sorry, I got distracted. Back now. Absolute value as a function either leaves the thing inside of it alone if it is 0 or positive, or multiplies it b negative 1 if it is negative. So |x|=x IF x is greater than or equal to 0, and |x| = -x IF x is less than 0. So for h(-2/3) you would have$h(\frac{-2}{3})=5\frac{|\frac{ - 2}{3}|}{\frac{-2}{3}}=5\frac{\frac{2}{3}}{\frac{-2}{3}}$

8. anonymous

Hmm, looks like the negative signs didn't display properly... Well, the 2s that are shifted way over to the right are supposed to have negative signs in front of them.

9. anonymous

so basictly anything negative inside the absolute value will turn into positive ?

10. anonymous

Yep!

11. anonymous

and once u turn the negative into positive then the absolute value sign goes away ?

12. anonymous

In this case, yes.

13. anonymous

14. anonymous

Ahh, well this depends on if a is positive or not. :P Find the answer for if a is positive, and then the answer for if a is negative. Then you say which one it is based on a.

15. anonymous

hmm ? what ? lol

16. anonymous

If $a \ge 0$then$h(3a)=5$and if$a < 0$then$h(3a)=-5$

17. anonymous

i see

18. anonymous

what do i have to do to know if a>0 ?

19. anonymous

how did you get h(3a)= 5 ? do you just cancel out the 3a ?

20. anonymous

Yeah.$5\frac{3a}{3a}=5$

21. anonymous

oh i see

22. anonymous

so if it was h(a-2) would the final answer be 5 anyway ?

23. anonymous

Either 5 or -5, depending on a.

24. anonymous

ok!

25. anonymous

gtg now, thanks again for helping me Aylin :)