`y = x^4 - 3x^3 + 3x^2 - x` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

Textbook Question

Chapter 4, Review - Problem 21 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The function has no vertical, horizontal or slant asymptotes.

You need to determine the extrema of the function, hence, you need to determine the zeroes of the first derivative:

`f'(x) = 4x^3-9x^2+6x-1 => f'(x) = 0 => (4x-1)(x-1)^2 = 0 => 4x -1 = 0 => x = 1/4`

`(x-1)^2 = 0 => x_1 = x_2 = 1`

Hence, the function has 2 extreme points, at `x = 1/4` and `x = 1.` Notice that `f'(x)<0` for `x in` `(-oo,1/4)` and `f'(x)>0` for `x in` `(1/4,oo)` , hence, the function decreases till `x = 1/4` and then it increases. Hence, the function has a minimum point at `x =1/4` .

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

`f''(x) = 12x^2 - 18x + 6 => f''(x) = 0 => 12x^2 - 18x + 6 = 0`

You need to divide by 6:

`2x^2 - 3x + 1 = 0 => x_(1,2) = (3+-sqrt(9-8))/4 => x_(1,2) = (3+-1)/4`

`x_1 = 1, x_2 = 1/2`

Hence, since the second derivative is negative on `(1/2,1) ` and positive on `(-oo,1/2) U (1,oo)` , then the function is concave down on `(1/2,1)` and concave up on `(-oo,1/2) U (1,oo).`

The graph of the function is represented below:

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