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Answer (i) : Given : f(x)= 1/(sqare root ( x  5 ) )
To find : range of f(x)
let y = f(x)
=> y = 1/ (sqare root(x5) )
squaring and inversing it, we get
=> 1/ y^{2} = x  5
=> x = ( 1/ y^{2} ) + 5
let x = g(y) = ( 1/ y^{2} ) + 5
=> range of f(x) = domain of g(y)
therefore , the domain of g(y) is R  {0 } which is range of f(x)
(ii) : Given : f(x)= 1/(sqare root (16  x ^{2}) )
To find : range of f(x)
let y = f(x)
=> y = 1/ (sqare root(16  x^{2}) )
squaring and inversing it, we get
=> 1/ y^{2} = 16 x^{2}
=> x^{2} = 16  ( 1/ y^{2} )
=> x = square root ( 16  ( 1/ y^{2} ) )
let x = g(y) = square root ( 16  ( 1/ y^{2} ) )
=> range of f(x) = domain of g(y)
therefore , the domain of g(y) is R  {0 } which is range of f(x)
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