What is the inverse of f(x)=-3(x-1)^2+2,x<= 1 ?

The inverse of f(x)=-3(x-1)^2+2,x<= 1 is sqrt(-(x-2)/3)+1,-sqrt(-(x-2)/3)+1

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inverse of f(x)=-3(x-1)^2+2,x<= 1

inverse

f (x) = - 3 (x - 1)^{2} + 2, x \leq 1

- \frac{x - 2}{3} + 1, - - \frac{x - 2}{3} + 1

Solution steps

y = - 3 (x - 1)^{2} + 2

Replace

x

with

y

x = - 3 (y - 1)^{2} + 2

Solve for

y, x = - 3 (y - 1)^{2} + 2

- \frac{x - 2}{3} + 1, - - \frac{x - 2}{3} + 1

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

f (x) = \frac{2 x - 1}{x ^{2} - 1}

f (x) = (x + 1) = 2 x

f (x) = 5^{x} - 3

f (x) = 9 + 3 x - 3

What is the inverse of f(x)=-3(x-1)^2+2,x<= 1 ?
The inverse of f(x)=-3(x-1)^2+2,x<= 1 is sqrt(-(x-2)/3)+1,-sqrt(-(x-2)/3)+1