Find the range of the function f(x)= x/3x^2+2 .

To find the codomain of the given function, we'll put f(x)
= y.

y=x/(3x^2+2)

We'll
multiply both sides by (3x^2+2):

y*(3x^2+2) =
x

We'll remove the
brackets:

3yx^2 + 2y = x

We'll
subtract x both sides:

3yx^2 + 2y - x =
0

We'll re-write the equation, ordering decreasingly the
powers of x:

3yx^2 - x + 2y =
0

This equation has real solutions if and only if the
discriminant delta > 0.

delta=b^2-4*a*c, where
a,b,c, are the coefficients of the quadratic,
ax^2+bx+c=0

delta=(-1)^2-24y^2

delta
= 1-24y^2

We'll consider the expresion of delta a
difference of squares:

delta =
(1-2ysqrt6)(1+2ysqrt6)

We'll solve the equation
delta=0.

(1-2ysqrt6)(1+2ysqrt6)=
0

We'll set each factor as
zero:

1-2ysqrt6 = 0

We'll
subtract 1 both sides:

-1 =
-2ysqrt6

We'll divide by
-2sqrt6:

y =
1/2sqrt6

y =
sqrt6/12

1+2ysqrt6 =
0

y=-sqrt6/12

Between
of the 2 roots, delta = 1-24y^2 is positive (because of the coefficient of y, which is
negative, -24).

So, the image of the function
is:

 Im f = [-sqrt6/12,
sqrt6/12]