B Blitze105 New member Joined Aug 28, 2008 Messages 27 Nov 16, 2008 #1 Hello i was just wondering if some one could check this for me. f(x)= x^3 - 3x^2 + 2 f'(x)= 3x^2 - 6x f''(x)= 6x - 6 Thus the point of inflection is (1,0) and from (-infinity, 1) is concave down and from (1,infinity) it is up
Hello i was just wondering if some one could check this for me. f(x)= x^3 - 3x^2 + 2 f'(x)= 3x^2 - 6x f''(x)= 6x - 6 Thus the point of inflection is (1,0) and from (-infinity, 1) is concave down and from (1,infinity) it is up
D Deleted member 4993 Guest Nov 16, 2008 #2 Blitze105 said: Hello i was just wondering if some one could check this for me. f(x)= x^3 - 3x^2 + 2 f'(x)= 3x^2 - 6x f''(x)= 6x - 6 Thus the point of inflection is (1,0) and from (-infinity, 1) is concave down and from (1,infinity) it is up Click to expand... Looks good to me. If it is not a computerized problem - sketch an approximate graph with your answer.
Blitze105 said: Hello i was just wondering if some one could check this for me. f(x)= x^3 - 3x^2 + 2 f'(x)= 3x^2 - 6x f''(x)= 6x - 6 Thus the point of inflection is (1,0) and from (-infinity, 1) is concave down and from (1,infinity) it is up Click to expand... Looks good to me. If it is not a computerized problem - sketch an approximate graph with your answer.
