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if ƒ:R→R be defined by ƒ(x) : x2+1 , then find ƒ-1(17) and ƒ-1(-3).
Please Give detailed solution.
The answer is given as :
ƒ-1(17) = {-4,4} , ƒ-1(-3) = ø
BASICALLY U NEED TO KNW THE VERY BASIC DEF OF INVERSE FN..
IF
Y=F(X)
X=F-1(Y)
NEEDLESS TO SAY THAT FOR AN EXAMPLE:
LEY Y=2*X X=3 => Y=6
SO, F-1(6)=3
UNDERSTOOD????
NOW F(X)=X^2+1
F-1(17)=X =>F(X)=17
SO X^22+1=17
SO X=+4 OR -4
SIMILARLY..........
X2+1=-3
=>X2=-4
NOT POSSIBLE
f(x) = x2 + 1
f(x) = y
x = f-1(y) .....................1
y = x2 + 1
(y-1)1/2 = x ..............2
from eq 1
f-1(y) = (y-1)1/2
so , f-1(x) = (x-1)1/2 , { for all x > = 1 }
f-117 = +(-)4 = {-4,4}
f-1-3 = @ , coz x cannot take any value less than 1 ...
f(x)= x2 + 1
f(1/x)= 1/(x2+1)
f-1(x)= 1/(x2+1)
f-1(17)= 1/(x2+1)
1/17 = 1/x2+1
17= x2 + 1
x2 = 16
x= {+4.-4}
Similarly for -3
x2 + 1 = -3
x2 = -4
x= under root -4
imaginary no.
hence f-1(-3) = 0.
plz approve....
let y= x2 + 1
=> x2 = y-1
=> x = √y-1
then replace y by x we get,
f-1 (x) = √x-1
then put the value of x.
Thanks a lot
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