What is the inverse of 7x^2-2,x>= 0 ?

The inverse of 7x^2-2,x>= 0 is sqrt((x+2)/7),-sqrt((x+2)/7)

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inverse of 7x^2-2,x>= 0

inverse

7 x^{2} - 2, x \geq 0

\frac{x + 2}{7}, - \frac{x + 2}{7}

Solution steps

y = 7 x^{2} - 2

Replace

x

with

y

x = 7 y^{2} - 2

Solve for

y, x = 7 y^{2} - 2

\frac{x + 2}{7}, - \frac{x + 2}{7}

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

(x - 6) (x - 8)

f (x) = \frac{4 x - 2}{x + 7}

\frac{- x + 1}{5 + 7 x}

1 - e^{- y}

What is the inverse of 7x^2-2,x>= 0 ?
The inverse of 7x^2-2,x>= 0 is sqrt((x+2)/7),-sqrt((x+2)/7)