My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree3Thanks
  • 2 Post By skeeter
Reply
 
LinkBack Thread Tools Display Modes
February 18th, 2019, 12:41 AM   #1
Newbie
 
MikeAndIke's Avatar
 
Joined: Feb 2019
From: N/A

Posts: 18
Thanks: 0

Math Focus: Calculus
Exclamation "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.
MikeAndIke is offline  
 
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.
skipjack is offline  
February 18th, 2019, 07:44 AM   #3
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,838
Thanks: 1481

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.
Thanks from skipjack and MikeAndIke
skeeter is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
derivatives, finding inflection points, inflection points, math, mikeandike, problem, problem solving, problems



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
inflection points analysis m.nikolov Calculus 2 January 12th, 2013 02:47 PM
inflection points cheyb93 Calculus 4 November 8th, 2012 11:30 AM
Finding Inflection Points soulrain Calculus 5 June 22nd, 2012 05:13 PM
inflection points questions, help? mathhelp123 Calculus 2 December 5th, 2009 07:34 AM
inflection points stainsoftime Calculus 3 December 15th, 2008 12:19 AM





Copyright © 2019 My Math Forum. All rights reserved.