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

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### 3 Answers

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}

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

Use y instead of f(x)

=>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**

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]**