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February 9th, 2018, 11:23 AM  #1 
Member Joined: Sep 2011 Posts: 98 Thanks: 1  differentiation / Integration Help
The curve has a gradient function dy/dx = 2 +q/(5x^2) where q is a constant, and a turning point at (0.5, 4). Find the value of q. option 1 : 2.5 option 2: 2.5 option 3: 0 Option 4: 3 I couldn't find the answer from any of the options and will need assistance to how the answer can be obtained. I have substituted x = 0.5 into dy/dx to get the gradient expression of 2 + 4q/5 and integrated to get y = 2x  q/(5x) + c. It seems impossible for me to get the value of q since c could not be found. I am not sure whether the question has some missing information to continue. Your help will be greatly appreciated. Thanks. Last edited by skipjack; February 9th, 2018 at 08:10 PM. 
February 9th, 2018, 11:28 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,550 Thanks: 2551 Math Focus: Mainly analysis and algebra 
What is the value of $\frac{\mathrm dy}{\mathrm dx}$ at a turning point?

February 9th, 2018, 05:34 PM  #3 
Member Joined: Sep 2011 Posts: 98 Thanks: 1 
It was not given to the value of dy/dx.

February 9th, 2018, 05:59 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 
a turning point is where the first derivative of a function changes sign $\dfrac{dy}{dx}=0$ at a turning point. $\dfrac{dy}{dx} = 2+\dfrac{q}{5x^2}$ at $(0.5, 4),~\dfrac{dy}{dx}=0$ $\left . 2+\dfrac{q}{5x^2} \right_{x=0.5} = 2+\dfrac{q}{5(0.5)^2} = 2+\dfrac{4q}{5} = 0$ $q = \dfrac 5 2 = 2.5$ i.e. choice 2 

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differentiation, ifferentiation, integration 
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