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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = \frac{3 x - 1}{4}

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

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

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

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