What is the inverse of f(x)=x^2-18bx ?

The inverse of f(x)=x^2-18bx is 9b+sqrt(x+81b^2),9b-sqrt(x+81b^2)

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inverse of f(x)=x^2-18bx

inverse

f (x) = x^{2} - 18 b x

9 b + x + 81 b^{2}, 9 b - x + 81 b^{2}

Solution steps

y = x^{2} - 18 b x

Replace

x

with

y

x = y^{2} - 18 b y

Solve for

y, x = y^{2} - 18 b y

9 b + x + 81 b^{2}, 9 b - x + 81 b^{2}

f (x) = 2 x^{2} - 8 x + 11

f (x) = \frac{e ^{x} - 4}{3 - e ^{x}}

f (x) = \frac{x + 2}{1 + x}

- cos (2 x)

What is the inverse of f(x)=x^2-18bx ?
The inverse of f(x)=x^2-18bx is 9b+sqrt(x+81b^2),9b-sqrt(x+81b^2)