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

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

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

inverse

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

2 x - 1, - 2 x - 1

Solution steps

y = (\frac{x}{2})^{2} + 1

Replace

x

with

y

x = (\frac{y}{2})^{2} + 1

Solve for

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

2 x - 1, - 2 x - 1

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

y = arctan (\frac{1}{x})

\frac{x - 7}{2 x + 1}

\frac{3 s + 6}{2 ( s + 0.43 ) ( s - 0.43 )}

sin (ln (\frac{π}{3}))

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