What is the inverse of f(x)=(t^2+3t-4)/(t^2-6t+5) ?

The inverse of f(x)=(t^2+3t-4)/(t^2-6t+5) is (4+5t)/(t-1)

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inverse of f(x)=(t^2+3t-4)/(t^2-6t+5)

inverse

f (x) = \frac{t ^{2} + 3 t - 4}{t ^{2} - 6 t + 5}

\frac{4 + 5 t}{t - 1}

Solution steps

x = \frac{t ^{2} + 3 t - 4}{t ^{2} - 6 t + 5}

Replace

t

with

x

t = \frac{x ^{2} + 3 x - 4}{x ^{2} - 6 x + 5}

Solve for

x, t = \frac{x ^{2} + 3 x - 4}{x ^{2} - 6 x + 5}

\frac{4 + 5 t}{t - 1}

f (x) = e (34 - 1)

x^{2} + 3 x + 9

x^{2} + 3 x + 4

(1 - 4 x)^{3}

What is the inverse of f(x)=(t^2+3t-4)/(t^2-6t+5) ?
The inverse of f(x)=(t^2+3t-4)/(t^2-6t+5) is (4+5t)/(t-1)