# How determine the range of function y=x/(x^4+3) in calculus?

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to determine the range of the given function, hence, you need to find the minimum and maximum values of function, using derivative of the function, such that:

`(dy)/(dx) = (d(x/(x^4+3)))/(dx)`

`(dy)/(dx) = ((x)'(x^4+3) - x*(x^4+3)')/((x^4+3)^2)`

`(dy)/(dx) = (x^4 + 3 - x*(4x^3))/((x^4+3)^2)`

`(dy)/(dx) = (x^4 + 3 - 4x^4)/((x^4+3)^2)`

`(dy)/(dx) = (3 - 3x^4)/((x^4+3)^2)`

Factoring out 3 yields:

`(dy)/(dx) = (3(1 - x^4))/((x^4+3)^2)`

Converting the difference of squares `1 - x^4 ` into a product, yields:

`(dy)/(dx) = (3(1 - x^2)(1 + x^2))/((x^4+3)^2)`

You may evaluate at what x values the function reaches its maximum and minimum values, solving the equation `(dy)/(dx) = 0` ,such that:

`3(1 - x^2)(1 + x^2) = 0`

Using zero product property yields:

`{(1 - x^2 = 0),(1 + x^2 != 0):}`

`x^2 = 1 => x_(1,2) = +-1`

You need to evaluate the minimum value `f(-1)` and the maximum value ` f(1)` , such that:

`f(-1) = -1/(1+3) => f(-1) = -1/4`

`f(1) = 1/4`

Hence, evaluating the range of the function, using the derivative of the function, yields `y in [-1/4,1/4].`