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# What are the local maxiumum and minimum points for the function f(x)=(3x)/(2x^2-18)

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What are the local maxiumum and minimum points for the function f(x)=(3x)/(2x^2-18)

Nov 27, 2018

#1
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As can be seen here : https://www.desmos.com/calculator/yhtjmyvehq

This function has no local max or min   Nov 27, 2018
#2
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I need to find it algebraically

Nov 27, 2018
#3
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I found the first derivative here : https://web2.0calc.com/questions/what-are-the-first-and-second-derivatives-of-the

as

-6 (x^2 + 9) (2x^2 - 18)^(-2)  .........set this to 0

-6 (x^2 + 9)

__________  =     0         multiply each side by (2x^2 - 18)   and we get

(2x^2 -18)^2

-6 (x^2 + 9)  =  0     divide both sides by - 6

x^2 + 9  =   0

Note that the left side is > 0  for all x

Thus....there are no critical points  ( max's or  mins )   Nov 27, 2018
#4
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Is the first derivative not -(6x^2+54)/(2x^2-18)^2 ?

Guest Nov 27, 2018
#5
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I could have taken the first derivative a little farther

-6 ( x^2 + 9)             -6(x^2 + 9)             -6 (x^2 + 9)           -3 (x^2 + 9)

___________ =     ____________ =    ____________ =   _________

(2x^2 - 18)^2         [ 2(x^2 - 9) ]^2         4 (x^2 - 9)^2          2 (x^2 - 9)

This is confirmed by WolframAlpha : https://www.wolframalpha.com/input/?i=derivative++(3x)+%2F+(2x%5E2+-+18)

Regardless.....there are still no max's or mins   CPhill  Nov 27, 2018