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
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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):
\[\pm \sqrt{x+36}\]
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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
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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"
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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