What is the inverse of (2s+1)/(3s^2+27) ?

The inverse of (2s+1)/(3s^2+27) is (1+sqrt(-3s(27s-1)+1))/(3s),(1-sqrt(-3s(27s-1)+1))/(3s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (2s+1)/(3s^2+27)

inverse

\frac{2 s + 1}{3 s ^{2} + 27}

\frac{1 + - 3 s ( 27 s - 1 ) + 1}{3 s}, \frac{1 - - 3 s ( 27 s - 1 ) + 1}{3 s}

Solution steps

y = \frac{2 s + 1}{3 s ^{2} + 27}

Replace

s

with

y

s = \frac{2 y + 1}{3 y ^{2} + 27}

Solve for

y, s = \frac{2 y + 1}{3 y ^{2} + 27}

\frac{1 + - 3 s ( 27 s - 1 ) + 1}{3 s}, \frac{1 - - 3 s ( 27 s - 1 ) + 1}{3 s}

f (x) = 0.95905

3 + 7 - 3 x^{2}

y = x^{2} - 1

f (x) = 2^{5 x}

What is the inverse of (2s+1)/(3s^2+27) ?
The inverse of (2s+1)/(3s^2+27) is (1+sqrt(-3s(27s-1)+1))/(3s),(1-sqrt(-3s(27s-1)+1))/(3s)