What is the inverse of f(x)= 1/4 x^2+x-1 ?

The inverse of f(x)= 1/4 x^2+x-1 is 2(sqrt(x+2)-1),-2(sqrt(x+2)+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)= 1/4 x^2+x-1

inverse

f (x) = \frac{1}{4} x^{2} + x - 1

2 (x + 2 - 1), - 2 (x + 2 + 1)

Solution steps

y = \frac{1}{4} x^{2} + x - 1

Replace

x

with

y

x = \frac{1}{4} y^{2} + y - 1

Solve for

y, x = \frac{1}{4} y^{2} + y - 1

2 (x + 2 - 1), - 2 (x + 2 + 1)

+ 2 x^{3}

f (x) = 500 + 35 x

f (x) = x^{2} + 4 x - 6, x \leq 2

f (x) = 5 \cdot x + \frac{π}{4}

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

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