What is the inverse of f(x)=2x^2-12x+13,x<= 1 ?

The inverse of f(x)=2x^2-12x+13,x<= 1 is (12+sqrt(8x+40))/4 ,(12-sqrt(8x+40))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=2x^2-12x+13,x<= 1

inverse

f (x) = 2 x^{2} - 12 x + 13, x \leq 1

\frac{12 + 8 x + 40}{4}, \frac{12 - 8 x + 40}{4}

Solution steps

y = 2 x^{2} - 12 x + 13

Replace

x

with

y

x = 2 y^{2} - 12 y + 13

Solve for

y, x = 2 y^{2} - 12 y + 13

\frac{12 + 8 x + 40}{4}, \frac{12 - 8 x + 40}{4}

y = 37.75 x + 3.64

f (x) = x^{3 + 4}

y = \frac{1}{16} x^{2}

f (x) = x^{3 + 1}

f (x) = x^{2} - 6 x, 3 \leq x \leq 5

What is the inverse of f(x)=2x^2-12x+13,x<= 1 ?
The inverse of f(x)=2x^2-12x+13,x<= 1 is (12+sqrt(8x+40))/4 ,(12-sqrt(8x+40))/4