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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

The range of the function is the set of all real values  of f(x) for the set of values x takes.

The denominator x^2+2  does not take 0.

Therefore for all values of x, we get a value of y.

Lt x--> -infinity   f(x) = Lt x /(3x^2+2) = 0

For all x < 0 f(x) = x/(3x^2+2) < 0 or negative as x.

Lt x--> infinity f(x) = x/(3x^2+2) = 0

Also for x > 0 ,  f(x) = x/(3x^2+2) > 0.

Further we see that f'(x) =   {(x)'(3x^2+2)-x(3x^2+2)}/(x^2+2)

f'(x) = {3x^2+2-6x^2}/(3x^2+2) =  (-3x^2+2)/(3x^2+2)

f'(x) = 0 gives 3x^2 = 2. Or x^2 =2/3 . So x = -sqrt2/3) and sqrt(2/3) are the extreme values .

Therefore  f(2/3) = (sqrt(2/3)/((2+2) = {sqrt(2/3)}/4 = (sqrt6)/12 and f(-2/3) = -(sqrt6)/12 are the extreme values of  f(x).

Therefore the range of  of f(x)  is { (-sqrt6)/12 , (sqrt6)/12}

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We have to find the range of f(x)= x/3x^2 + 2

Range is all the values that f(x) can take for all possible values of x.

Now f(x)= x/3x^2 + 2

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

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

=> 3yx^2 + 2y = x

=> 3yx^2 + 2y - x = 0

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

Now  3yx^2 - x + 2y = 0 has real solutions if (-1)^2 - 4*3y*2 >=0

=> 1 - 24 y >=0

=> (1 - 2y sqrt 6)(1 + 2y sqrt 6) > =0

This gives either (1 - 2y sqrt 6) and (1 + 2y sqrt 6) >= 0

or (1 - 2y sqrt 6) and (1 + 2y sqrt 6) =< 0

From the two we get that the range is all values between -sqrt6/12 and sqrt6/12 inclusive.

Therefore the range is all values of x between -sqrt 6 / 12 and sqrt 6 / 12 inclusive

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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]