What is the inverse of f(x)=(x^2+1)/3 ?

The inverse of f(x)=(x^2+1)/3 is sqrt(3x-1),-sqrt(3x-1)

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inverse of f(x)=(x^2+1)/3

inverse

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

3 x - 1, - 3 x - 1

Solution steps

y = \frac{x ^{2} + 1}{3}

Replace

x

with

y

x = \frac{y ^{2} + 1}{3}

Solve for

y, x = \frac{y ^{2} + 1}{3}

3 x - 1, - 3 x - 1

f (x) = - 0.009 x^{2} + 0.392 x - 0.0259

f (x) = 3^{x - 2} - 4

0.5 cos (6 x)

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

5 ln (2 x - 7)

What is the inverse of f(x)=(x^2+1)/3 ?
The inverse of f(x)=(x^2+1)/3 is sqrt(3x-1),-sqrt(3x-1)