What is the inverse of (x^2+1)/(2x^2+7) ?

The inverse of (x^2+1)/(2x^2+7) is sqrt((1-7x)/(2x-1)),-sqrt((1-7x)/(2x-1))

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inverse of (x^2+1)/(2x^2+7)

inverse

\frac{x ^{2} + 1}{2 x ^{2} + 7}

\frac{1 - 7 x}{2 x - 1}, - \frac{1 - 7 x}{2 x - 1}

Solution steps

y = \frac{x ^{2} + 1}{2 x ^{2} + 7}

Replace

x

with

y

x = \frac{y ^{2} + 1}{2 y ^{2} + 7}

Solve for

y, x = \frac{y ^{2} + 1}{2 y ^{2} + 7}

\frac{1 - 7 x}{2 x - 1}, - \frac{1 - 7 x}{2 x - 1}

2 lo g_{4} (5 x + 1)

f (x) = 0.5 (3 - 3 x)

f (x) = \frac{3}{x - 1} - 3

f (x) = \frac{x - 5}{x + 6}

\frac{1}{s ^{5}}

What is the inverse of (x^2+1)/(2x^2+7) ?
The inverse of (x^2+1)/(2x^2+7) is sqrt((1-7x)/(2x-1)),-sqrt((1-7x)/(2x-1))