A curve is defined for x 0 and is such that dy/dx = x + 4/x^2. The point P(4,8) lies on the curve. Show that the gradient of the curve has a minimum value when x = 2 and state this minimum value.

Question

A curve is defined for x > 0 and is such that dy/dx = x + 4/x^2. The point P(4,8) lies on the curve. Show that the gradient of the curve has a minimum value when x = 2 and state this minimum value.

psymon · Answer

My brain must be missing something and I wish I had more time to figure it out. What you can do to get the y-value for x = 2 is integrate then use the point (4,8) to help find your C. Then use that function to find y when x = 2. My brain isn't working for the show part.

lncognlto · Answer

Ok, so I've worked it out and the integration comes out at y = x^2/2 - 4/x + 1, and y = 1 when x = 2...

praxer · Answer

Find the first and second derivative and then equate the first derivative to 0 then place the root of the equation in the second derivative equation. If it is positive then the minima is obtained at that point and if negative then maxima is obtained. The points where the minima is obtained gives the minimum value and the maximum point will give the maximum value.

amistre64 · Answer

A curve is defined for x > 0 
     such that dy/dx = x + 4/x^2. 

* integrate it


The point P(4,8) lies on the curve. 

* apply this initial condition


Show that the gradient of the curve has 
       a minimum value when x = 2 
           and state this minimum value.

*should be self explanatory afterwards

amistre64 · Answer

dy/dx is already defined for you, so determine its critical points to test for min/max with

0s and undefineds

lncognlto · Answer

Ok, thanks, guys!