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

Guest Nov 27, 2018
 #1
avatar+92641 
+1

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

 

This function has no local max or min

 

cool cool cool

CPhill  Nov 27, 2018
 #2
avatar
+1

I need to find it algebraically 

Guest Nov 27, 2018
 #3
avatar+92641 
+1

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 )

 

 

cool cool cool

CPhill  Nov 27, 2018
 #4
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0

Is the first derivative not -(6x^2+54)/(2x^2-18)^2 ?

Guest Nov 27, 2018
 #5
avatar+92641 
+1

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

 

 

cool cool cool

CPhill  Nov 27, 2018

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