What is the inverse of 2(x+1)^2-5 ?

The inverse of 2(x+1)^2-5 is sqrt((x+5)/2)-1,-sqrt((x+5)/2)-1

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inverse of 2(x+1)^2-5

inverse

2 (x + 1)^{2} - 5

\frac{x + 5}{2} - 1, - \frac{x + 5}{2} - 1

Solution steps

y = 2 (x + 1)^{2} - 5

Replace

x

with

y

x = 2 (y + 1)^{2} - 5

Solve for

y, x = 2 (y + 1)^{2} - 5

\frac{x + 5}{2} - 1, - \frac{x + 5}{2} - 1

x^{3} - 48 x

f (x) = \frac{x - 2}{x ^{2} + 2 x - 15}

\frac{3 x + 4}{x ^{2} - 25}

y = 5 x + 2, at x = 1

f (x) = \frac{2}{x - 6}

What is the inverse of 2(x+1)^2-5 ?
The inverse of 2(x+1)^2-5 is sqrt((x+5)/2)-1,-sqrt((x+5)/2)-1