What is the inverse of f(x)=x^2+3.5x+6 ?

The inverse of f(x)=x^2+3.5x+6 is (-7+sqrt(16x-47))/4 ,-(sqrt(16x-47)+7)/4

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inverse of f(x)=x^2+3.5x+6

inverse

f (x) = x^{2} + 3.5 x + 6

\frac{- 7 + 16 x - 47}{4}, - \frac{16 x - 47 + 7}{4}

Solution steps

y = x^{2} + 3.5 x + 6

Replace

x

with

y

x = y^{2} + 3.5 y + 6

Solve for

y, x = y^{2} + 3.5 y + 6

\frac{- 7 + 16 x - 47}{4}, - \frac{16 x - 47 + 7}{4}

f (x) = 3 x + y = 3

\frac{1}{2} sin (x - \frac{π}{2})

f (x) = x^{3} + x^{2} - 16 x - 16

f (x) = 2 x^{2} - 4

\frac{x}{2} + \frac{y}{5} - 1 = 0

What is the inverse of f(x)=x^2+3.5x+6 ?
The inverse of f(x)=x^2+3.5x+6 is (-7+sqrt(16x-47))/4 ,-(sqrt(16x-47)+7)/4