Question

graphing the sq rt. functions:
g(x)=sq rt x-1 (x=0,1,4,9)
i understand to replace x with the numbers to come up with 4 relations, but my problem is after subtracting the 1 under the sq rt, my problem is removing the sq rt symbol, yes i know th sq the number, but how can you sq a (0,1,3)

Answer 1

you can take only the approximate values

Answer 2

so do i quit the problem and write "function cannot be defined"?

Answer 3

for my answer

Answer 4

my book has answers to this

Answer 5

wouldn't you take for x =(1,5,10,..) ?

Answer 6

hang on, i took a picture from my book, let me figure out how to get it to u

Answer 7

ok

Answer 8

unless u wanna give me ur # and i can just text to u, if not hang on

Answer 9

look at the bottom, however it is the same problem

Answer 10

√4-1=1.....how?
√0-1=-1...how?

Answer 11

oh, I thought it was \[\sqrt{x-1}\]
but it's
\[\sqrt x-1\]
so for x=4 you have
\[\sqrt4 -1=2-1=1\]
for x =0:
\[\sqrt0 -1=0-1=-1\]
and so on

Answer 12

ooooh, only first # is under the sq rt?

Answer 13

yes

Answer 14

i feeeeeel stttuuuuuuuupppppppiiiiiiidddddd

Answer 15

I've been stupid so many times, but this motivates you to become better

Answer 16

f(x)=x^4-x^2+1 x=-x
when replacing is it
x^4-(-x)+1 or x^4-x+1

Answer 17

\[(-x)^4 -(-x)^2+1\]

Answer 18

but you should have put it on new post

Answer 19

y u got - in front of x^4?

Answer 20

I replaced x with -x

Answer 21

original dont have it

Answer 22

no on first variable, maybe u made mistake

Answer 23

f(x)=x^4-x^2+......see

Answer 24

but (-x)^4 is simply x^4

Answer 25

ooh shucks, sorry, im trippy mane!

Answer 26

new post time but it picks up from here!