What is the inverse of f(x)=x^2+bx+1 ?

The inverse of f(x)=x^2+bx+1 is (-b+sqrt(b^2-4(1-x)))/2 ,(-b-sqrt(b^2-4(1-x)))/2

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inverse of f(x)=x^2+bx+1

inverse

f (x) = x^{2} + b x + 1

\frac{- b + b ^{2} - 4 ( 1 - x )}{2}, \frac{- b - b ^{2} - 4 ( 1 - x )}{2}

Solution steps

y = x^{2} + b x + 1

Replace

x

with

y

x = y^{2} + b y + 1

Solve for

y, x = y^{2} + b y + 1

\frac{- b + b ^{2} - 4 ( 1 - x )}{2}, \frac{- b - b ^{2} - 4 ( 1 - x )}{2}

f (x) = \frac{5 x + 2}{x - 9}

f (x) = \frac{3 - x}{7}

f (x) = \frac{3}{x + 5}

e^{x} - 5

(\frac{x ^{2}}{4} - \frac{x}{2} - \frac{1}{4})^{2}

What is the inverse of f(x)=x^2+bx+1 ?
The inverse of f(x)=x^2+bx+1 is (-b+sqrt(b^2-4(1-x)))/2 ,(-b-sqrt(b^2-4(1-x)))/2