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]

x2 + 1  =  10

x2  =  10 - 1

x2  =  9

x  =  ± 3

x2 + 1  =  37

x2  =  37 - 1

x2  =  36

x  =  ± 6

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

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