What is the inverse of f(x)=(-2/3)x^2+(1/3) ?

The inverse of f(x)=(-2/3)x^2+(1/3) is sqrt(-(3x-1)/2),-sqrt(-(3x-1)/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of f(x)=(-2/3)x^2+(1/3)

inverse

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

- \frac{3 x - 1}{2}, - - \frac{3 x - 1}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{3 x - 1}{2}, - - \frac{3 x - 1}{2}

What is the inverse of f(x)=(-2/3)x^2+(1/3) ?
The inverse of f(x)=(-2/3)x^2+(1/3) is sqrt(-(3x-1)/2),-sqrt(-(3x-1)/2)