f(x) = one million / (x + 3) + 3 isn't its very own inverse although, f(x) = one million / (x + 3) - 3 is its very own inverse one thank you to tutor this: f(f^-one million(x)) = x f( f(x)) = one million / [(one million / (x + 3) - 3) + 3] - 3 = x? f(f(x)) = one million / [one million / (x + 3)] - 3 f(f(x)) = (x + 3) - 3 = x tests so particular, the composition of f(x) and f(x) = x, showing that f is its very own inverse yet another approach: locate the inverse of f(x) write as y = one million / (x + 3 ) - 3 change x and y x = one million / (y + 3) - 3 now resolve for y x + 3 = one million / (y + 3) (x + 3)(y + 3) = one million y + 3 = one million / (x + 3) y = one million / (x + 3) - 3 = f^-one million(x) so f(x) = f^-one million(x), confirming that f(x) is its very own inverse
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f(3) = 9 * 3 + 1 = 27 + 1 = 28
I'm not sure what you mean by f-1? Do you mean f^(-1), the inverse of f? If so, then it's 3. The inverse of a function reverses the function.
f(x) = one million / (x + 3) + 3 isn't its very own inverse although, f(x) = one million / (x + 3) - 3 is its very own inverse one thank you to tutor this: f(f^-one million(x)) = x f( f(x)) = one million / [(one million / (x + 3) - 3) + 3] - 3 = x? f(f(x)) = one million / [one million / (x + 3)] - 3 f(f(x)) = (x + 3) - 3 = x tests so particular, the composition of f(x) and f(x) = x, showing that f is its very own inverse yet another approach: locate the inverse of f(x) write as y = one million / (x + 3 ) - 3 change x and y x = one million / (y + 3) - 3 now resolve for y x + 3 = one million / (y + 3) (x + 3)(y + 3) = one million y + 3 = one million / (x + 3) y = one million / (x + 3) - 3 = f^-one million(x) so f(x) = f^-one million(x), confirming that f(x) is its very own inverse