impossible stationary points (1 Viewer)

[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.

 

[OkDen]

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.

 

[cos16mh]

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

 

[fan96]

Its the second derivative that I'm struggling with

Then test the first derivative around the turning points.