What is the inverse of (k^2+16)/(x^2+1) ?

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

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inverse of (k^2+16)/(x^2+1)

inverse

\frac{k ^{2} + 16}{x ^{2} + 1}

\frac{k ^{2} + 16 - x}{x}, - \frac{k ^{2} + 16 - x}{x}

Solution steps

y = \frac{k ^{2} + 16}{x ^{2} + 1}

Replace

x

with

y

x = \frac{k ^{2} + 16}{y ^{2} + 1}

Solve for

y, x = \frac{k ^{2} + 16}{y ^{2} + 1}

\frac{k ^{2} + 16 - x}{x}, - \frac{k ^{2} + 16 - x}{x}

f (x) = \frac{8 x - 3}{5 x + 1}

f (x) = sin (x + 5)

\frac{s - 6}{s ^{2} - 2 s + 10}

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

\frac{5}{4 + x ^{2}}

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