What is the inverse of f(x)= 1/(4pi^2)x(4pi-x) ?

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

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inverse of f(x)= 1/(4pi^2)x(4pi-x)

inverse

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

- 2 π (- x + 1 - 1), 2 π (- x + 1 + 1)

Solution steps

y = \frac{1}{4 π ^{2}} x (4 π - x)

Replace

x

with

y

x = \frac{1}{4 π ^{2}} y (4 π - y)

Solve for

y, x = \frac{1}{4 π ^{2}} y (4 π - y)

- 2 π (- x + 1 - 1), 2 π (- x + 1 + 1)

(0, 0), (1, 2)

f (x) = \frac{1 - x ^{2}}{x}

2 (x - 3) + 5

f (x) = \frac{e ^{x}}{1 - e ^{x}}

- 3 x^{3} + 5 x^{2} + 16 x

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