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:
No comments:
Post a Comment