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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

x^{2} + 12 x + 36

y = 2 x^{2} - 8

\frac{x ^{3} - 9 x}{10}

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

(0, 0), (- 3, 5)

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