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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{9}{t}

f (x) = \frac{1}{x ^{2} - 7 x - 8}

f (x) = \frac{( 2 x - 6 )}{( x ^{(2)} + 4 x - 5 )}

y = - 2 x - 3

f (x) = \frac{3 x ^{2} - 108}{x ^{2} - 6 x}

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