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February 18th, 2019, 12:41 AM  #1 
Newbie Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus  "Finding Inflection Points" Math Problems Help
I am just learning this new topic at school, and was hoping that you can help me with this example my teacher provided me. The problem is: Let f(x)=2x^48x^3+180x^2. For what values of x does the graph of f have a point of inflection? I understand that you have to do something with the second derivative but I don't remember how to start off the entire process, could anyone please help in explaining and showing each step individually from beginning to end? The problem I had to solve for homework was: Let f(x)=0.5x^46x^327x^2. For what values of x does the graph of f have a point of inflection? The second derivative I got was:6x^236x^154 which equals 6(x+3)(x+3)=0, so the value of x would just be x=3 right? Last edited by skipjack; February 18th, 2019 at 02:48 PM. 
February 18th, 2019, 04:48 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,648 Thanks: 2085 
The first derivative of f(x) is 8x³  24x² + 360x, so its second derivative is 24x²  48x + 360, which equals 24(x + 5)(x  3). Hence the second derivative is zero for x = 5 and for x = 3. The first derivative has a turning point for these values of x, so they're the values of x for which f(x) has a point of inflection. For the homework problem, yes, there's just one inflection point, where x = 3. Correction (thanks, skeeter): x = 3 isn't a point of inflection. The first derivative has a point of inflection, not a turning point, at x = 3. Last edited by skipjack; February 18th, 2019 at 03:01 PM. 
February 18th, 2019, 07:44 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,925 Thanks: 1521  Quote:
$f''(x) = 6(x+3)^2 = 0$ at $x=3$ is true, but f''(x) does not change sign there.  

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