FINDING THE INVERSE OF AN ELEMENT IF THE FUNCTION IS GIVEN

Question 1 :

If f -> Q -> Q is given by f(x)  =  x², then find

(i)  f^-1 (9)  (ii)  f^-1(-5)  (iii)  f^-1(0)

Solution :

From the given function, we have

 f(x)  =  x²

x  =  f^-1(x²)  ----(1)

(i)  f^-1 (9)

By comparing f^-1 (9) with (1), we get that 

x²  =  9

x  =  ±3

(ii)  f^-1(-5)  

By comparing f^-1 (5) with (1), we get that 

x²  = -5

x  =  √-5 which is not possible

f^-1(-5)  is undefined.

(iii)  f^-1(0)

By comparing f^-1 (0) with (1), we get that 

x²  =  0

x  =  0

Hence the value of f^-1(0)   =  0.

Question 2 :

In the function f : R-> R be defined by f(x)  =  x² + 5x + 9, find f^-1 (8) and f^-1(9)

Solution :

Given that :  f(x)  =  x² + 5x + 9

x  =  f^-1(x² + 5x + 9)  ---(1)

f^-1 (8)  ---(2)

By comparing (1) and (2), we have 8 instead of x² + 5x + 9

x² + 5x + 9  =  8

x² + 5x + 9 - 8  =  0

x² + 5x + 1  =  0

x  =  (-b±√b² - 4ac)/2a

x  =  (-5±√(5²-4))/2(1)

x  =  (-5±√21)/2

f^-1(9)

x² + 5x + 9  =  9

x² + 5x + 9 - 9  =  0

x² + 5x  =  0

x(x + 5)  =  0

x  =  0 and x  =  -5

Question 3 :

Let R-> R be defined on f(x)  =  x² + 1. Find 

(i)  f^-1 (-5)  (ii)  f^-1(26)  (iii)  f^-1[10, 37]

Solution :

f(x)  =  x² + 1

(i)  f^-1 (-5)  

x² + 1  =  -5

x²  =  -6 Which is not real.

(ii)  f^-1(26)

x² + 1  =  26

x²  =  26 - 1

x²  =  25

x  =  ± 5

(iii)  f^-1[10, 37]

Hence the values of x are [-6, -3, 3, 6].

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