What is the inverse of f(x)=(3x)/((sqrt(7x^2+16))) ?

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

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inverse of f(x)=(3x)/((sqrt(7x^2+16)))

inverse

f (x) = \frac{3 x}{( 7 x ^{2} + 16 )}

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

Solution steps

y = \frac{3 x}{7 x ^{2} + 16}

Replace

x

with

y

x = \frac{3 y}{7 y ^{2} + 16}

Solve for

y, x = \frac{3 y}{7 y ^{2} + 16}

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

y = (e)^{x}

1 (0.5)^{x}

e^{- \frac{x ^{4}}{2}}

f (x) = (- 1, 2), (4, 1)

f (x) = 5 - x + 6

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