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July 16th, 2017, 03:55 PM   #1
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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.
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July 16th, 2017, 04:07 PM   #2
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Any thoughts? Do you know how to find inflection points?
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July 16th, 2017, 04:07 PM   #3
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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
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July 16th, 2017, 04:44 PM   #4
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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?
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July 16th, 2017, 04:51 PM   #5
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Quote:
Originally Posted by mathishard23 View Post
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 $x$ ...
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July 16th, 2017, 05:53 PM   #6
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x=6 and x=5/16 ?
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July 16th, 2017, 06:10 PM   #7
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Quote:
Originally Posted by mathishard23 View Post
x=6 and x=5/16 ?
set each factor equal to zero and solve

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

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

try again
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July 16th, 2017, 08:49 PM   #8
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Quote:
Originally Posted by skeeter View Post
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?
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July 16th, 2017, 11:21 PM   #9
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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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