What is the inverse of h(x)=(4x-1)/(3x+7) ?

The inverse of h(x)=(4x-1)/(3x+7) is (-1-7x)/(3x-4)

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inverse of h(x)=(4x-1)/(3x+7)

inverse

h (x) = \frac{4 x - 1}{3 x + 7}

\frac{- 1 - 7 x}{3 x - 4}

Solution steps

y = \frac{4 x - 1}{3 x + 7}

Replace

x

with

y

x = \frac{4 y - 1}{3 y + 7}

Solve for

y, x = \frac{4 y - 1}{3 y + 7}

\frac{- 1 - 7 x}{3 x - 4}

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

x^{4} + x^{2} + 3

f (x) = \frac{1 + 9 x}{4 - x}

y = \frac{e ^{x}}{1 + 8 e ^{x}}

f (x) = ln (16 x + e)

What is the inverse of h(x)=(4x-1)/(3x+7) ?
The inverse of h(x)=(4x-1)/(3x+7) is (-1-7x)/(3x-4)