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

The inverse of-(x+1)^2-2,(x<= 1) is sqrt(-x-2)-1,-sqrt(-x-2)-1

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inverse of-(x+1)^2-2,(x<= 1)

inverse

- (x + 1)^{2} - 2, (x \leq 1)

- x - 2 - 1, - - x - 2 - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

- x - 2 - 1, - - x - 2 - 1

sin (0.9622)

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

y = \frac{x ^{2} - 6 x - 7}{x + 5}

\frac{( 9 x - 4 )}{4 x + 1}

\frac{1}{x ^{2} + 4 x + 4}

What is the inverse of-(x+1)^2-2,(x<= 1) ?
The inverse of-(x+1)^2-2,(x<= 1) is sqrt(-x-2)-1,-sqrt(-x-2)-1