# Determine the image of the function y=x/(3x^2+2)

hala718 | Certified Educator

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

Cross multiply:

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

==> 3yx^2 + 2y - x = 0

==> (3y)x^2 - x + 2y = 0

==> x1 = [1 + sqrt(1-4*3y*2y)]/2*3y

= [1+ sqrt(1-24y^2)]/6y

We notice that `the only option we have in order to have a REAL solution is when (1-24y^2 = 0   , otherwise we will have negative square root and no REAL solution.

==> 1-24y^2 = 0

==> 24y^2 = 1

==> y^2 = 1/24

==> y= +- 1/sqrt24 = +- 1/2sqrt6

Then Im = [ -1/2sqrt6, 1/2sqrt6]

neela | Student

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

Here We transform the denomonator  3x^2+2 = t.

x = Then 6x dx = dt . Or xdx = dt/6.

So the given integral  now becomes:

Int y dx = Int  {xdx/(3x^2+2) } =  Int dt/6t = (1/6) dt/t = (1/6)lnt

Int Ydx = (1/6)ln (3x^2+2).

giorgiana1976 | Student

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]