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 February 9th, 2018, 10:23 AM #1 Member   Joined: Sep 2011 Posts: 97 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 07:10 PM.
 February 9th, 2018, 10:28 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra What is the value of $\frac{\mathrm dy}{\mathrm dx}$ at a turning point?
 February 9th, 2018, 04:34 PM #3 Member   Joined: Sep 2011 Posts: 97 Thanks: 1 It was not given to the value of dy/dx.
 February 9th, 2018, 04:59 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,981 Thanks: 1027 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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