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 July 16th, 2017, 03:55 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 second-derivative 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 second-derivative 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, 04:07 PM #2 Senior Member   Joined: Oct 2009 Posts: 142 Thanks: 60 Any thoughts? Do you know how to find inflection points?
 July 16th, 2017, 04:07 PM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,662 Thanks: 1327 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, 04:44 PM #4 Newbie   Joined: May 2017 From: antartica Posts: 11 Thanks: 0 so for question one, i dound that the second derivative is -96x^2-30x and when setting that to equal 0 I got -30/96 =x. Is that correct?
July 16th, 2017, 04:51 PM   #5
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Quote:
 Originally Posted by mathishard23 so for question one, i dound that the second derivative is -96x^2-30x and when setting that to equal 0 I got -30/96 =x. Is that correct?
Quote:
 Test f(x)=−8x^4−5x^3+7 for concavity and inflection points.
$f'(x) = -32x^3 - 15x^2$

$f''(x) = -96x^2 - 30x = -6x(16x + 5) = 0$

there are two solutions for $x$ ...

 July 16th, 2017, 05:53 PM #6 Newbie   Joined: May 2017 From: antartica Posts: 11 Thanks: 0 x=6 and x=5/16 ?
July 16th, 2017, 06:10 PM   #7
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Quote:
 Originally Posted by mathishard23 x=6 and x=5/16 ?
set each factor equal to zero and solve

$-6x=0$ ... $x=?$

$16x+5=0$ ... $x=?$

try again

July 16th, 2017, 08:49 PM   #8
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Quote:
 Originally Posted by skeeter set each factor equal to zero and solve $-6x=0$ ... $x=?$ $16x+5=0$ ... $x=?$ try again
isnt 0/6 just 0 and x=-5/16?

 July 16th, 2017, 11:21 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.

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