What is the inverse of f(x)=3(x+2)^2-5,x<=-2 ?

The inverse of f(x)=3(x+2)^2-5,x<=-2 is sqrt((x+5)/3)-2,-sqrt((x+5)/3)-2

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inverse of f(x)=3(x+2)^2-5,x<=-2

inverse

f (x) = 3 (x + 2)^{2} - 5, x \leq - 2

\frac{x + 5}{3} - 2, - \frac{x + 5}{3} - 2

Solution steps

y = 3 (x + 2)^{2} - 5

Replace

x

with

y

x = 3 (y + 2)^{2} - 5

Solve for

y, x = 3 (y + 2)^{2} - 5

\frac{x + 5}{3} - 2, - \frac{x + 5}{3} - 2

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

f (x) = 4 x - 9

f (x) = \frac{5 x - 2}{4 - 5 x}

f (x) = \frac{x}{9 + x ^{2}}

f (x) = 4 - 7 x

What is the inverse of f(x)=3(x+2)^2-5,x<=-2 ?
The inverse of f(x)=3(x+2)^2-5,x<=-2 is sqrt((x+5)/3)-2,-sqrt((x+5)/3)-2