What is the inverse of f(x)=b_{1}x^2 ?

The inverse of f(x)=b_{1}x^2 is sqrt(x/(b_{1))},-sqrt(x/(b_{1))}

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inverse of f(x)=b_{1}x^2

inverse

f (x) = b_{1} x^{2}

\frac{x}{b _{1}}, - \frac{x}{b _{1}}

Solution steps

y = b_{1} x^{2}

Replace

x

with

y

x = b_{1} y^{2}

Solve for

y, x = b_{1} y^{2}

\frac{x}{b _{1}}, - \frac{x}{b _{1}}

f (x) = - x

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

(4 x - 1)^{(- 1)} [(4 (- 2) - 1)]

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

y = \frac{2 x - 1}{3}

What is the inverse of f(x)=b_{1}x^2 ?
The inverse of f(x)=b_{1}x^2 is sqrt(x/(b_{1))},-sqrt(x/(b_{1))}