What is the inverse of f(x)=(10x)/(x^2+49) ?

The inverse of f(x)=(10x)/(x^2+49) is (5+sqrt(-49x^2+25))/x ,(5-sqrt(-49x^2+25))/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=(10x)/(x^2+49)

inverse

f (x) = \frac{10 x}{x ^{2} + 49}

\frac{5 + - 49 x ^{2} + 25}{x}, \frac{5 - - 49 x ^{2} + 25}{x}

Solution steps

y = \frac{10 x}{x ^{2} + 49}

Replace

x

with

y

x = \frac{10 y}{y ^{2} + 49}

Solve for

y, x = \frac{10 y}{y ^{2} + 49}

\frac{5 + - 49 x ^{2} + 25}{x}, \frac{5 - - 49 x ^{2} + 25}{x}

f (x) = 6 x - 4 y = 24

(- 1.8) (4.3)

6 x + 3 y = 6

f (x) = - 2 x + 9

5 x + 3 y = 15

What is the inverse of f(x)=(10x)/(x^2+49) ?
The inverse of f(x)=(10x)/(x^2+49) is (5+sqrt(-49x^2+25))/x ,(5-sqrt(-49x^2+25))/x