Find F^-1 (x) when F(x)=3x^2 +1?

Let f (x) = y = 3x² + 1

3x² = y - 1
x² = (y - 1) / 3
x = √ [ (y - 1) / 3 ]
g(y) = √ [ (y - 1) / 3 ]

f^(-1) (x) = √ [ (x - 1) / 3 ]

it's just the function of x to the power of minus 1.

F^-1 (x) = 1/ 3x^2 +1

f(x) = 3x^2 + 1

Change f(x) into y:
f(x) = 3x^2 + 1
y = 3x^2 + 1
3x^2 = y - 1
x^2 = (y - 1)/3
x = √[(y - 1)/3]

Change x into f^-1(x) and y into x:
x = √[(y - 1)/3]
f^-1(x) = √[(x - 1)/3]

1 Change F(x) with y
2 Change y whit x, and solve for y

F(x)=3x^2 +1
y=3x^2 +1
x=3y^2 +1
x-1=3y^2
x-1/3=y^2
SQUR(x-1/3) = y

So F^-1 (x) = SQUR(x-1/3)

set it up as

y= 3x^2 +1

Now swap all your x's and y's

x=3y^2 +1

y^2 = (x-1)/3

y= +/- ((x-1)/3)^.5

gotta make sure your domain and range match up