What is the inverse of f(x)=4x^2+1,x>= 0 ?

The inverse of f(x)=4x^2+1,x>= 0 is (sqrt(x-1))/2 ,-(sqrt(x-1))/2

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inverse of f(x)=4x^2+1,x>= 0

inverse

f (x) = 4 x^{2} + 1, x \geq 0

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

Solution steps

y = 4 x^{2} + 1

Replace

x

with

y

x = 4 y^{2} + 1

Solve for

y, x = 4 y^{2} + 1

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

- x^{2} - 4 x + 4

f (x) = 10 x^{3} + 3

\frac{2 x}{x - 1}

f (x) = 6.5 - 12 x

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

What is the inverse of f(x)=4x^2+1,x>= 0 ?
The inverse of f(x)=4x^2+1,x>= 0 is (sqrt(x-1))/2 ,-(sqrt(x-1))/2