Alexis87

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.

v8archie

What is the value of $\frac{\mathrm dy}{\mathrm dx}$ at a turning point?

Alexis87

It was not given to the value of dy/dx.

romsek

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