Question

h(3)
h(x)= 5|x|/x

Answer 1

@Aylin

Answer 2

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

Answer 3

That looks right.

Answer 4

alright cool :D

Answer 5

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 ?

Answer 6

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

Answer 7

@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\[h(\frac{-2}{3})=5\frac{|\frac{ - 2}{3}|}{\frac{-2}{3}}=5\frac{\frac{2}{3}}{\frac{-2}{3}}\]

Answer 8

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.

Answer 9

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

Answer 10

Yep!

Answer 11

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

Answer 12

In this case, yes.

Answer 13

what about h(3a) ?

Answer 14

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.

Answer 15

hmm ? what ? lol

Answer 16

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

Answer 17

i see

Answer 18

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

Answer 19

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

Answer 20

Yeah.\[5\frac{3a}{3a}=5\]

Answer 21

oh i see

Answer 22

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

Answer 23

Either 5 or -5, depending on a.

Answer 24

ok!

Answer 25

gtg now, thanks again for helping me Aylin :)