What is the inverse of f(x)=(x^2)/4-x-1 ?

The inverse of f(x)=(x^2)/4-x-1 is 2(sqrt(x+2)+1),2(-sqrt(x+2)+1)

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inverse of f(x)=(x^2)/4-x-1

inverse

f (x) = \frac{x ^{2}}{4} - x - 1

2 (x + 2 + 1), 2 (- x + 2 + 1)

Solution steps

y = \frac{x ^{2}}{4} - x - 1

Replace

x

with

y

x = \frac{y ^{2}}{4} - y - 1

Solve for

y, x = \frac{y ^{2}}{4} - y - 1

2 (x + 2 + 1), 2 (- x + 2 + 1)

f (x) = 11 x^{2} + 2

f (x) = 11 x^{2} + 1

e^{(2 x + 1)}

2 x + 4 + x^{2}, - 3 < x < 2

f (x) = 3 sin (x) - 6

What is the inverse of f(x)=(x^2)/4-x-1 ?
The inverse of f(x)=(x^2)/4-x-1 is 2(sqrt(x+2)+1),2(-sqrt(x+2)+1)