 # 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 12 years ago

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

12 years ago

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 ...

12 years ago

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....

12 years ago

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.

12 years ago

Thanks a lot