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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^4-8x^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^4-6x^3-27x^2. For what values of x does the graph of f have a point of inflection? The second derivative I got was:-6x^2-36x^1-54 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: 20,969 Thanks: 2219 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
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Quote:
 The problem I had to solve for homework was: Let f(x)=-0.5X^4-6X^3-27X^2. For what values of x does the graph of f have a point of inflection? The second derivative I got was:-6x^2-36x^1-54 which equals -6(x+3)(x+3)=0, so the value of x would just be x=-3 right?
The posted homework function has no inflection points. An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes; i.e. the sign of the second derivative changes from positive to negative or negative to positive.

$f''(x) = -6(x+3)^2 = 0$ at $x=-3$ is true, but f''(x) does not change sign there. Tags derivatives, finding inflection points, inflection points, math, mikeandike, problem, problem solving, problems Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post m.nikolov Calculus 2 January 12th, 2013 02:47 PM cheyb93 Calculus 4 November 8th, 2012 11:30 AM soulrain Calculus 5 June 22nd, 2012 05:13 PM mathhelp123 Calculus 2 December 5th, 2009 07:34 AM stainsoftime Calculus 3 December 15th, 2008 12:19 AM

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