We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
106
5
avatar

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

 Nov 27, 2018
 #1
avatar+101360 
+1

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

 

This function has no local max or min

 

cool cool cool

 Nov 27, 2018
 #2
avatar
+1

I need to find it algebraically 

 Nov 27, 2018
 #3
avatar+101360 
+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

 Nov 27, 2018
 #4
avatar
0

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

Guest Nov 27, 2018
 #5
avatar+101360 
+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

17 Online Users

avatar
avatar
avatar