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What is image of function y = x /(1+4x^2)?

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greenbel | (Level 2) Honors

Posted June 25, 2013 at 1:38 PM via web

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What is image of function y = x /(1+4x^2)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 25, 2013 at 2:19 PM (Answer #1)

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The image of the function is the result returned by the function `f(x) = y` for each x input from domain of the function.

You should start by moving all terms to one side, such that:

`x/(1+4x^2) - y = 0`

Bringing the terms to a common denominator yields:

`x - y*(1 + 4x^2) = 0`

`x - y - 4x^2y = 0`

You need to re-arrange the terms such that:

`- 4x^2y + x - y = 0`

You need to notice that the equation `- 4x^2y + x - y = 0` has real solutions if its discriminant `Delta = b^2 - 4ac >= 0` .

You need to identify the coefficients a,b,c such that:

`a = -4y, b = 1, c = -y`

`Delta = 1^2 - 4*(-4y)*(-y)`

`Delta = 1^2 - 16y^2`

Converting the difference of squares into a product yields:

`Delta = (1 - 4y)(1 + 4y)`

You need to solve for y the inequality `Delta >= 0` such that:

`(1 - 4y)(1 + 4y) = 0 => {(1 - 4y = 0),(1 + 4y = 0):}`

`{(y = 1/4),(y = -1/4):}`

You need to notice that `Delta >= 0` for `y in [-1/4,1/4]` .

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

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