What is the inverse of 7/(9x^2-16) ?

The inverse of 7/(9x^2-16) is (sqrt(x)sqrt(7+16x))/(3x),-(sqrt(x)sqrt(7+16x))/(3x)

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inverse of 7/(9x^2-16)

inverse

\frac{7}{9 x ^{2} - 16}

\frac{x 7 + 16 x}{3 x}, - \frac{x 7 + 16 x}{3 x}

Solution steps

y = \frac{7}{9 x ^{2} - 16}

Replace

x

with

y

x = \frac{7}{9 y ^{2} - 16}

Solve for

y, x = \frac{7}{9 y ^{2} - 16}

\frac{x 7 + 16 x}{3 x}, - \frac{x 7 + 16 x}{3 x}

x^{2} + x + 1

f (x) = \frac{x ^{2} + 4}{x - 3}

f (x) = \frac{x + 7}{x + 4}

f (x) = 6 x - 12 - 3 x^{2}

y = 12 \cdot \frac{( 3 ) ( 2 )}{4}

What is the inverse of 7/(9x^2-16) ?
The inverse of 7/(9x^2-16) is (sqrt(x)sqrt(7+16x))/(3x),-(sqrt(x)sqrt(7+16x))/(3x)