What is the inverse of ((t-1)^2)/3 ?

The inverse of ((t-1)^2)/3 is sqrt(3)sqrt(t)+1,-sqrt(3)sqrt(t)+1

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inverse of ((t-1)^2)/3

inverse

\frac{( t - 1 ) ^{2}}{3}

3 t + 1, - 3 t + 1

Solution steps

y = \frac{( t - 1 ) ^{2}}{3}

Replace

t

with

y

t = \frac{( y - 1 ) ^{2}}{3}

Solve for

y, t = \frac{( y - 1 ) ^{2}}{3}

3 t + 1, - 3 t + 1

f (x) = \frac{7 x - 1}{3 x + 2}

f (x) = 2 x + 1 f (x) = x^{2} + 4

f (x) = x + 6 + 1

f (x) = \frac{8 x}{5 x - 4}

f (x) = - 5 x - 3 + 4

What is the inverse of ((t-1)^2)/3 ?
The inverse of ((t-1)^2)/3 is sqrt(3)sqrt(t)+1,-sqrt(3)sqrt(t)+1