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

The inverse of f(x)= 1/(2(x-4)^2+1) is (16x+sqrt(-8x^2+8x))/(4x),(16x-sqrt(-8x^2+8x))/(4x)

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

inverse

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

\frac{16 x + - 8 x ^{2} + 8 x}{4 x}, \frac{16 x - - 8 x ^{2} + 8 x}{4 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{16 x + - 8 x ^{2} + 8 x}{4 x}, \frac{16 x - - 8 x ^{2} + 8 x}{4 x}

(x - 1)^{3} + 5

f (x) = 300 - 50 x

\frac{x - 2}{2 x - 3}

f (x) = \frac{5 x + 1}{2 x - 3}

What is the inverse of f(x)= 1/(2(x-4)^2+1) ?
The inverse of f(x)= 1/(2(x-4)^2+1) is (16x+sqrt(-8x^2+8x))/(4x),(16x-sqrt(-8x^2+8x))/(4x)