Question

g(x) =(ax+b) /(cx+d), here ad is not equal to bc. Then domain of g^-1

Answer 1

we have to find domain of g¯¹ when g(x) = (ax + b)/(cx + d), here ad ≠ bc.

solution : let y = g(x) = (ax + b)/(cx + d)

⇒y(cx + d) = (ax + b)

⇒cxy + dy = ax + b

⇒cxy - ax = b - dy

⇒x(cy - a) = (b - dy)

⇒x = (a - dy)/(cy - a)

⇒f(y) = (a - dy)/(cy - a)

now putting x in place of y,

⇒f(x) = (a - dx)/(cx - a)

here f(x) = inverse of g(x) i.e., g¯¹

so domain of f(x) = domain of g¯¹ = R - {a/c}

Therefore the domain of g¯¹ is R - {a/c}.

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