`y = 2 - 2x - x^3` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

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Chapter 4, Review - Problem 19 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the horizontal asymptotes to the graph of function, such that:

`lim_(x->+oo) 2-2x-x^3 = oo`

`lim_(x->-oo) 2-2x-x^3 = oo`

Notice that the graph has no vertical or horizontal asymptotes. You need to check if there are slant asymtpotes, y = ax+b, such that:

`a = lim_(x->oo) (f(x))/x = lim_(x->oo) (2-2x-x^3)/x = oo`

Since a ->oo, yields that there are no slant asymptotes.

You need to determine the extremes of the functions, hence, you need to evaluate the zeroes of derivative of function, such that:

`f'(x) = -2 - 3x^2 =>f'(x) = 0=> -2 - 3x^2 = 0=> 3x^2 = -2 => x^2 = -2/3 => x_(1,2) !in R`

Hence, there are no absolute extrema of the function.

You need to determine the inflection points, hence, you need to find the zeroes of the second derivative:

`f''(x) = -6x => f''(x) = 0 => -6x = 0 => x = 0 => f(0) = 2.`

Hence, the function has an inflection point (0,2).

The graph of the function `f(x) = 2-2x-x^3` , whose inflection point,where the graph changes its curvature, is at (0,2), is represented below.

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