What is the inverse of f(x)=3(x+1)^2-4 ?

The inverse of f(x)=3(x+1)^2-4 is sqrt((x+4)/3)-1,-sqrt((x+4)/3)-1

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inverse of f(x)=3(x+1)^2-4

inverse

f (x) = 3 (x + 1)^{2} - 4

\frac{x + 4}{3} - 1, - \frac{x + 4}{3} - 1

Solution steps

y = 3 (x + 1)^{2} - 4

Replace

x

with

y

x = 3 (y + 1)^{2} - 4

Solve for

y, x = 3 (y + 1)^{2} - 4

\frac{x + 4}{3} - 1, - \frac{x + 4}{3} - 1

y = 2 x^{2} - 3

2 x + 6

x^{2} + 4 x + 7

\frac{x}{( x - 1 ) ( x + 5 )}

5 x + 3 y = 6

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