What is the inverse of f(x)=5x^2+2,x>= 0 ?

The inverse of f(x)=5x^2+2,x>= 0 is sqrt((x-2)/5),-sqrt((x-2)/5)

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inverse of f(x)=5x^2+2,x>= 0

inverse

f (x) = 5 x^{2} + 2, x \geq 0

\frac{x - 2}{5}, - \frac{x - 2}{5}

Solution steps

y = 5 x^{2} + 2

Replace

x

with

y

x = 5 y^{2} + 2

Solve for

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

\frac{x - 2}{5}, - \frac{x - 2}{5}

y = 0.5 x + 2

f (x) = \frac{5 x + 1}{- x + 7}, x \neq = 7

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

f (x) = \frac{x + 4}{2 x - 3}

What is the inverse of f(x)=5x^2+2,x>= 0 ?
The inverse of f(x)=5x^2+2,x>= 0 is sqrt((x-2)/5),-sqrt((x-2)/5)