Question

Metal g^-1(x)
Show the steps it takes to arrive at the answer (x+1)^1/2
If I'm late to respond its because of my computer

Answer 1

I don't get it. How can we arrive at (x+1)^(1/2) if we aren't given anything?
Is metal a math terminology? If so what does it mean to metal g^(-1)(x)?

Or maybe g(x)=(x+1)^(1/2) and you want to find the g^(-1)(x)?

Answer 2

\[y=(x+1)^\frac{1}{2}\]
square both sides first
and solve for x

Answer 3

AHhhh, brain fart! f(x)=2x+1 and g(x)=x^2-1

Answer 4

we only need g(x)=x^2-1 to find g^(-1)(x)

Answer 5

y=x^2-1

add 1 on both sides as a first step

Answer 6

you are trying to solve for x

Answer 7

please show what you have after doing that step

Answer 8

y+(1)=x^2-1+(1) is this properly set up?

Answer 9

y+1=x^2 correct

Answer 10

now you can take the square root of both sides !
if you assume x>0 you will have your answer

Answer 11

wouldn't the problem look like y+1=x^2= y+1/x=x? where is the 1/2 coming into play? Sorry for my ignorance I just haven't done these in some time

Answer 12

I don't know what you did to get all of that

Answer 13

\[y+1=x^2\]
I know to solve for x you take square root of both sides

and if you assume x>0 you have your answer

Answer 14

you do know \[\sqrt{9}=\sqrt{3^2}=3\]
you see that the square root and the square cancel right?

Answer 15

Yes

Answer 16

Example below:
just like if you are solving
\[x^2=9 \\ \text{ you would take square root of both sides } \\ x=\sqrt{9}=3 \text{ or } x=-\sqrt{9}=-3\]

Answer 17

here you do the same thing
\[x^2=y+1\]
\[\text{ if } x>0 \text{ then } x=\sqrt{y+1} \\ \\ \text{ if } x<0 \text{ then } x=-\sqrt{y+1}\]

Answer 18

now you interchange the variables
and after doing that you can call y g^(-1)(x) if you choose

Answer 19

I see. (1+x)^1/2 is the same as sqrt x+1

Answer 20

yes

Answer 21

danke