What is the inverse of-5x^2+x,x>= 1 ?

The inverse of-5x^2+x,x>= 1 is -(-1+sqrt(1-20x))/(10),-(-1-sqrt(1-20x))/(10)

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inverse of-5x^2+x,x>= 1

inverse

- 5 x^{2} + x, x \geq 1

- \frac{- 1 + 1 - 20 x}{10}, - \frac{- 1 - 1 - 20 x}{10}

Solution steps

y = - 5 x^{2} + x

Replace

x

with

y

x = - 5 y^{2} + y

Solve for

y, x = - 5 y^{2} + y

- \frac{- 1 + 1 - 20 x}{10}, - \frac{- 1 - 1 - 20 x}{10}

f (x) = 2 e^{x - 1}

f (x) = \frac{7 x + 1}{- 4 x + 9}

f (x) = 0.008 x^{3}

g (x) = - 5 x + 20

f (x) = \frac{6 x + 1}{x - 5}

What is the inverse of-5x^2+x,x>= 1 ?
The inverse of-5x^2+x,x>= 1 is -(-1+sqrt(1-20x))/(10),-(-1-sqrt(1-20x))/(10)