impossible stationary points
Hi I would like some assistance on this question Thanks
find any turning point on the curve y=(x-3)√(4-x) and determine their nature.
Re: impossible stationary points
Quote:
Originally Posted by
cos16mh
Hi I would like some assistance on this question Thanks
find any turning point on the curve y=(x-3)√(4-x) and determine their nature.
For Turning Points, dy/dx = 0
To find derivative of y=(x-3)√(4-x) use Product Rule.
So, dy/dx = (-3x+11)/2√(-x+4)
Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)
Which gives: x = 11/3
Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9
To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.
positive = minimum
negative = maximium
Or you can graph it, or test coordinates.
Re: impossible stationary points
Quote:
Originally Posted by
OkDen
For Turning Points, dy/dx = 0
To find derivative of y=(x-3)√(4-x) use Product Rule.
So, dy/dx = (-3x+11)/2√(-x+4)
Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)
Which gives: x = 11/3
Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9
To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.
positive = minimum
negative = maximium
Or you can graph it, or test coordinates.
Its the second derivative that I'm struggling with
Re: impossible stationary points
Quote:
Originally Posted by
cos16mh
Its the second derivative that I'm struggling with
Then test the first derivative around the turning points.