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impossible stationary points (1 Viewer)

cos16mh

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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

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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

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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
 

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