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

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

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

inverse

f (x) = 4 x^{2} + 1

\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}

f (x) = \frac{3 x + 4}{5 - 4 x}

x - 4 = 0

f (x) = x^{3} - 1

f (x) = 3 x^{2} - 3

f (x) = x^{4} - 8 x + 16

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