solution for the given problems

Asked by sankarbongu2003 | 22nd Jun, 2020, 11:18: AM

Expert Answer:

To find the domain of f(x) where it is given as
straight f left parenthesis straight x right parenthesis equals fraction numerator straight x minus 3 over denominator open parentheses straight x plus 3 close parentheses square root of straight x squared minus 4 end root end fraction
straight f left parenthesis straight x right parenthesis space is space defined space if space open parentheses straight x plus 3 close parentheses square root of straight x squared minus 4 end root not equal to 0
straight x not equal to negative 3 space and space square root of straight x squared minus 4 end root not equal to 0
square root of straight x squared minus 4 end root space is space defined space when space straight x squared minus 4 greater than 0
rightwards double arrow left parenthesis straight x minus 2 right parenthesis left parenthesis straight x plus 2 right parenthesis greater than 0
rightwards double arrow straight x minus 2 greater than 0 space and space straight x plus 2 greater than 0 space space OR space space straight x minus 2 less than 0 space and space straight x plus 2 less than 0
rightwards double arrow straight x greater than 2 space and space straight x greater than negative 2 space space OR space space straight x less than 2 space and space straight x less than negative 2
rightwards double arrow straight x greater than 2 space space OR space space straight x less than negative 2
rightwards double arrow straight x element of left parenthesis 2 comma space infinity right parenthesis space space OR space space straight x element of left parenthesis negative infinity comma space minus 2 right parenthesis
rightwards double arrow straight x element of left parenthesis negative infinity comma space minus 2 right parenthesis union left parenthesis 2 comma space infinity right parenthesis
Hence comma space the space domain space of space straight f left parenthesis straight x right parenthesis space is space left parenthesis negative infinity comma space minus 2 right parenthesis union left parenthesis 2 comma space infinity right parenthesis minus left curly bracket 3 right curly bracket

Answered by Renu Varma | 22nd Jun, 2020, 12:01: PM