# answering the problem. y= (x^3)/(x+1)^2 Following will be considered when answering the problem. y= (x^3)/(x+1)^2 a) Find the equation to maximized or minimized. b) Finding the solution c) Showing that your solution is an absolute maximized or minimized. d) Sketch the curve e) Find an equation of the slant asymptote. a) You need to remember that the equation of derivative is the equation you need to use when the problem requests an optimization.

You need to differentiate the function with respect to x such that:

` y' = ((x^3)'*(x+1)^2 - (x^3)*((x+1)^2)')/((x+1)^4)`

`y' = ((3x^2)*(x+1)^2 - 2(x^3)*(x+1))/((x+1)^4)`

`y' = (x^2(x+1)(3x + 3 - 2x))/((x+1)^4)`

`y' = (x^2(x + 3))/((x+1)^3)`

Hence, evaluation the equation that needs to be optimized yields `y' = (x^2(x + 3))/((x+1)^3).`

b) You need to solve the equation y'=0 such that:

`(x^2(x + 3))/((x+1)^3) = 0 =gt (x^2(x + 3)) = 0`

`x^2 = 0 =gt x = 0`

`x + 3 = 0 =gt x = -3`

Hence, evaluating the solutions to equation y'=0 yields x = 0 and x = -3.

c) You need to select a value for x between -3 and 0 such that:

`x = -2 =gt y'(-2) = 4(-2 + 3)/((-2+1)^3) = -4`

Hence, the function decreases over (-3,0) and increases over `(-oo,-3)`  and `(0,oo).`

Thus, the function reaches its absolute maximum at x=-3 and its absolute minimum at x=0.

d)

e) You need to remember the equation of slant asymptote such that:

`y = mx + b`

`m = lim_(x-gt+-oo)f(x)/x`

`lim_(x-gt+-oo)(x^3)/(x(x+1)^2) = lim_(x-gt+-oo)(x^2)/((x+1)^2) =1`

Since m is finite, then you may evaluate b such that:

`b = lim_(x-gt+-oo)(f(x) - mx)`

`b = lim_(x-gt+-oo)((x^3)/((x+1)^2) - x)`

`b = lim_(x-gt+-oo)((x^3 - x^3 - 2x^2 - x)/((x+1)^2)) = -2`

Hence, evaluating the equation of the slant asymptote yields:`y = x - 2.`

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