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Mathematics 16 Online
OpenStudy (princesssleelee):

Which function below is the inverse of f(x) = x2 − 36? x squared over 36 ±6square root of x 1 over quantity x squared minus 36 ±square root of the quantity x plus 36

OpenStudy (md152727):

f(x) = x^2 - 36 y = x^2 - 36 .... replace f(x) with y x = y^2 - 36 .... swap x and y now solve for y to get the inverse

OpenStudy (princesssleelee):

solve for y=x^2 -36, correct?

OpenStudy (md152727):

yes

OpenStudy (princesssleelee):

Being option D?

OpenStudy (princesssleelee):

\[\pm \sqrt{x+36}\]

OpenStudy (md152727):

Nope

OpenStudy (md152727):

f(x) = x^2 - 36 y = x^2 - 36 .... replace f(x) with y x = y^2 - 36 .... swap x and y now solve for y to get the inverse

OpenStudy (md152727):

f(x)=y=x2−36inverse⟹x=y2−36 notice to get the "inverse relation", we'd firsts swap about the variables, then solve for "y"

OpenStudy (princesssleelee):

I tried that, I cant get it.

OpenStudy (md152727):

f(x)=y=x2−36inverse⟹x=y2−36x+36=y2then take −−−−√ to both sides

OpenStudy (md152727):

consider f(x) = y y = x^2 - 36 y+36 = x^2 x = sqrt(y+36) now consider f(x) = y => x =f^-1(y) => f^-1(y) = sqrt(y+36) therefore f^-1(x) = sqrt(x+36) Did you get it now???

OpenStudy (md152727):

? :?

OpenStudy (pawanyadav):

Yes,, D is right

OpenStudy (princesssleelee):

I'm sorry, no. I get a complete mind block.

OpenStudy (princesssleelee):

@Pawanyadav How did you conclude "D"

OpenStudy (pawanyadav):

As you already have y+36=x^2 So for inverse interchange x,y You will get x+36=y^2 Square root both side +-√(x+36)=y so D

OpenStudy (princesssleelee):

Okay, thank you.

OpenStudy (pawanyadav):

You're welcome

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