What is the inverse of f(x)=2x^2-8x,x>= 2 ?

The inverse of f(x)=2x^2-8x,x>= 2 is (8+sqrt(64+8x))/4 ,(8-sqrt(64+8x))/4

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inverse of f(x)=2x^2-8x,x>= 2

inverse

f (x) = 2 x^{2} - 8 x, x \geq 2

\frac{8 + 64 + 8 x}{4}, \frac{8 - 64 + 8 x}{4}

Solution steps

y = 2 x^{2} - 8 x

Replace

x

with

y

x = 2 y^{2} - 8 y

Solve for

y, x = 2 y^{2} - 8 y

\frac{8 + 64 + 8 x}{4}, \frac{8 - 64 + 8 x}{4}

\frac{4 + x}{2 - x}

f (x) = \frac{- 3 x + 15}{x ^{2} - 5 x}

f (x) = \frac{x ^{2} - 4 x + 6}{( x - 2 ) ^{2}}

f (x) = 8 x

ln (3) e^{x}

What is the inverse of f(x)=2x^2-8x,x>= 2 ?
The inverse of f(x)=2x^2-8x,x>= 2 is (8+sqrt(64+8x))/4 ,(8-sqrt(64+8x))/4