Question

Find its inverse
(a) g(x)=(x-1)/(3x+2)
(b) f(x)=4+ square root of (x^2+1)

Answer 1

$(a)\\ g(x)=\dfrac{(x-1)}{(3x+2)} \implies (3x+2)g(x)=x-1\\ \implies3xg(x)+2g(x)-x+1=0 \implies 2g(x)+1=x-3xg(x) \\ \implies x=\dfrac{1+2g(x)}{1-3g(x)} \\ \\\implies\ g^{-1}(x)=\dfrac{1+2x}{1-3x}; x{=}\mathllap{/\,}1/3\\ \\ (b) \\ f(x)=4+\sqrt{\smash[b]{x^2+1}}; R(f):f\ge5 \implies f(x)-4=\sqrt{x^2+1}\\ \implies (f(x)-4)^2=x^2+1\\ \implies (f(x)-4)^2-1=x^2\\ \implies x=\sqrt{(f(x)-4)^2-1}\\ f^{-1}(x)=\sqrt{{(x-4)^2-1}}; x\ge5$