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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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].`

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