What is the inverse of f(x)=16x^2+1 ?

The inverse of f(x)=16x^2+1 is (sqrt(x-1))/4 ,-(sqrt(x-1))/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)=16x^2+1

inverse

f (x) = 16 x^{2} + 1

\frac{x - 1}{4}, - \frac{x - 1}{4}

Solution steps

y = 16 x^{2} + 1

Replace

x

with

y

x = 16 y^{2} + 1

Solve for

y, x = 16 y^{2} + 1

\frac{x - 1}{4}, - \frac{x - 1}{4}

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

- 2 x^{2} - 2 x - 2

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

f (x) = \frac{x + 20}{x - 5}

(- 4, - 2), (6, 2)

What is the inverse of f(x)=16x^2+1 ?
The inverse of f(x)=16x^2+1 is (sqrt(x-1))/4 ,-(sqrt(x-1))/4