My Math Forum "Finding Inflection Points" Math Problems Help

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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: 20,373 Thanks: 2010 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.

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