Question

math problem can someone check my answer please :)

Answer 1

@NormaValenzuela

Answer 2

@Ignaris

Answer 3

@triciaal

Answer 4

\(\normalsize\color{blueviolet}{ \rm f(x)=\sqrt{x-5} }\)

replace \(\normalsize\color{blueviolet}{ \rm f(x) }\) with \(\normalsize\color{blueviolet}{ \rm y }\).

\(\normalsize\color{blueviolet}{ \rm y=\sqrt{x-5} }\)

replace \(\normalsize\color{blueviolet}{ \rm y }\) and \(\normalsize\color{blueviolet}{ \rm x }\) with each other:

\(\normalsize\color{blueviolet}{ \rm x=\sqrt{y-5} }\)

Answer 5

so wit would be A?

Answer 6

then you need to solve for y.

(The restriction that you choose, that \(\normalsize\color{blueviolet}{ \rm y\ge5 }\) is correct, because if \(\normalsize\color{blueviolet}{ \rm y<5 }\), then you would end up with a negative in the square root.)

Answer 7

\(\normalsize\color{blueviolet}{ \rm x=\sqrt{y-5} }\)

raise both sides to the second power:

\(\normalsize\color{blueviolet}{ \rm (x)^2=( \sqrt{y-5}~)^2 }\)

Answer 8

what do you get?

Answer 9

oh B

Answer 10

yes

Answer 11

thanks :)

Answer 12

For the function f(x) = x2 - 12, find (f o f-1)(4)

Answer 13

would this be 0

Answer 14

@SolomonZelman wouldn't this be 4?

Answer 15

ohh its four?

Answer 16

@DanJS

Answer 17

@zappy not sure

Answer 18

:(

Answer 19

Oh, I thought you are correcting me.... as if I said anything, anyways:P
what exactly are you looking for though?

Answer 20

For the function f(x) = x2 - 12, find (f o f-1)(4)

Answer 21

is it 0

Answer 22

what is meant by (f o f -1)(4)
you are trying to say, \(\normalsize\color{royalblue}{ \rm f(f^{-1}(4)) }\) ?

Answer 23

so its 4

Answer 24

(original) \(\normalsize\color{royalblue}{ \rm f(x)=\sqrt{x-5} }\)
(inverse) \(\normalsize\color{royalblue}{ \rm f^{-1}(x)=x^2+5 }\) \(\huge\color{white}{ \left| \right| }\)

\(\normalsize\color{royalblue}{ \rm f(~\color{red}{f^{-1}(\color{blue}{x})}~)=\sqrt{\color{red}{\color{blue}{x}^2+5}-5} }\)

\(\normalsize\color{royalblue}{ \rm f(~\color{red}{f^{-1}(\color{blue}{x})}~)=\sqrt{\color{red}{\color{blue}{x}^2}} }\)

\(\normalsize\color{royalblue}{ \rm f(~\color{red}{f^{-1}(\color{blue}{x})}~)=x }\)

Answer 25

next step, is on you. plug in 4 for x.

Answer 26

and yes, it is 4.

Answer 27

ok thanks :)