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Consider `f: R->R`given by `f(x) = 4x + 3`. Show that `f` is invertible. Find the inverse of `f`. |
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Answer» `f: R → R` is given by, `f(x) = 4x + 3` Let `f(x) = f(y)` `=>4x+3 = 4y+3` `=>4x=4y` `=>x = y` `:. f` is a one-one function. For `y in R`, Let `y = 4x+3` `:. x = (y-3)/4` Now, `f(x) = f((y-3)/4) = 4((y-3)/4)+3 = y` `:. f` is an onto function. Let `g:R->R` such that `g(x) = (y-3)/4` Then, `gof(x) = g(f(x)) = ((4x+3)-3 )/4 = x` `fog(y) = f(g(y)) = 4((y-3)/4)+3 = y` `:. gof = fog = I_R` Hence, `f` is invertible and the inverse of f is given by `f^-1(y)=g(y)=(y-3)/4`. |
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