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

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

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inverse of f(x)=(x+1)^2+5

inverse

f (x) = (x + 1)^{2} + 5

x - 5 - 1, - x - 5 - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

x - 5 - 1, - x - 5 - 1

\frac{s + 3}{( s + 1 ) ( s + 4 )}

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

\frac{2 s + 3}{s ^{2} + 9}

h (t) = - 16 t^{2} + 30 t + 5

f (x) = 1.25 (38 - x)

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