
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
July 16th, 2017, 02:55 PM  #1 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0  Curve Sketching Concavity questions
1. Test f(x)=−8x^4−5x^3+7 for concavity and inflection points. Solution: f has inflection points when x=____ and x=____. (List the smaller value first.) f is concave down on (−infinity,_____) U (_______, infinity) and its concave up on (_____,_______). 2. Test the function g(x)=(−8x^3) − (−132)x^2 + (−576)x− 5 for relative maximum and minimum. Use the secondderivative test if possible. Solution: From g'(x)= Drop down menu with following options (pick one) [(8x^2)(132)x +(576)], [(24x^2)(132)x + (576)], [(24x^2)(264)x + (576)] it follows that the critical values of g are x1= ____and x2=_____ . (List the smaller value first.) Since g''(x)= drop down menu [(48x)(132)], [(48x)(264)] we have that g"(x1)=_____ when ( >0 OR <0) and g"(x2)= ________ when ( >0 OR <0) Therefore, by the secondderivative test, there is a (relative minimum OR relative maximum) when x=x1. and there is a (relative minimum OR relative maximum) when x=x2. Thanks so much. 
July 16th, 2017, 03:07 PM  #2 
Senior Member Joined: Oct 2009 Posts: 275 Thanks: 92 
Any thoughts? Do you know how to find inflection points?

July 16th, 2017, 03:07 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376 
1) for any polynomial function, inflection points occur where f''(x) changes sign ... that would mean you would first have to find where f''(x) = 0 2) a polynomial function has a maximum where f'(x) = 0 and f''(x) < 0 ... a minimum where f'(x) = 0 and f''(x) > 0 
July 16th, 2017, 03:44 PM  #4 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0 
so for question one, i dound that the second derivative is 96x^230x and when setting that to equal 0 I got 30/96 =x. Is that correct?

July 16th, 2017, 03:51 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376  Quote:
Quote:
$f''(x) = 96x^2  30x = 6x(16x + 5) = 0$ there are two solutions for $x$ ...  
July 16th, 2017, 04:53 PM  #6 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0 
x=6 and x=5/16 ?

July 16th, 2017, 05:10 PM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376  
July 16th, 2017, 07:49 PM  #8 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0  
July 16th, 2017, 10:21 PM  #9 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0 
So... I got it thanks for all the help. For the second equation just find the first derivate and its x values there will be 2 of them. Then find the second derivaive and take your x1 and x2 and plug those in. You should get 2 numbers and one will be greater than 0 and the other less than 0.


Tags 
concavity, curve, questions, sketching 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sketching a curve  axerexa  Calculus  1  April 1st, 2015 08:29 AM 
curve sketching  Lima  Calculus  1  April 1st, 2013 08:08 PM 
Curve Sketching Help!  ProJO  Calculus  3  March 27th, 2011 08:50 AM 
Curve Sketching  hunnybee  Calculus  7  March 22nd, 2011 10:55 PM 
Curve Sketching: Concavity HELP PLEASE!  ProJO  Calculus  2  March 20th, 2011 02:23 PM 