A Arkan New member Joined Feb 28, 2015 Messages 1 Feb 28, 2015 #1 Of this function: f(x) = x^2(1-x) This is my work: Determining second dericative: y = x^2-x^3 y' = 2x - 3x^2 y'' = 2-6x Determining concavity: 2-6x>0 -6x>-2 x<1/3 But when I look at the graph it looks like the answer should be x<2/3
Of this function: f(x) = x^2(1-x) This is my work: Determining second dericative: y = x^2-x^3 y' = 2x - 3x^2 y'' = 2-6x Determining concavity: 2-6x>0 -6x>-2 x<1/3 But when I look at the graph it looks like the answer should be x<2/3
I Ishuda Elite Member Joined Jul 30, 2014 Messages 3,342 Feb 28, 2015 #2 Arkan said: Of this function: f(x) = x^2(1-x) This is my work: Determining second dericative: y = x^2-x^3 y' = 2x - 3x^2 y'' = 2-6x Determining concavity: 2-6x>0 -6x>-2 x<1/3 But when I look at the graph it looks like the answer should be x<2/3 Click to expand... The answer x\(\displaystyle \le\)1/3 is correct for the graph being concave. A local maximum occurs at x=2/3 (and a local minimum at x=0)
Arkan said: Of this function: f(x) = x^2(1-x) This is my work: Determining second dericative: y = x^2-x^3 y' = 2x - 3x^2 y'' = 2-6x Determining concavity: 2-6x>0 -6x>-2 x<1/3 But when I look at the graph it looks like the answer should be x<2/3 Click to expand... The answer x\(\displaystyle \le\)1/3 is correct for the graph being concave. A local maximum occurs at x=2/3 (and a local minimum at x=0)
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,250 Mar 1, 2015 #3 Ishuda said: The answer x\(\displaystyle \le\)1/3 is correct for the graph being concave [up]. Click to expand... The answer is x < 1/3.
Ishuda said: The answer x\(\displaystyle \le\)1/3 is correct for the graph being concave [up]. Click to expand... The answer is x < 1/3.