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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

sin (x) + 8

f (x) = \frac{x + 1}{x - 2} = 10

(0, 4), (2, - 2)

8 x^{4} - 48 x^{2}

f (x) = x - 7

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