y = ((x - 1)^3)/(2*x)
Q01. Diagram
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Q02. y = ((x - 1)^4)/(2*x)
Properties of the curve
- Sign of x and y
- x < 0 and y > 0 : Curve in 2nd quadrant
- x between 0 and 1 and y < 0 : Curve in 4th quadrant
- x = 1 and y = 0
- x > 1 and y > 0 and from 0 to +infinite : Curve in 1st quadrant
- Asymptote
- x = 0 and y = infinite. Hence x = 0 is vertical asymptote
- x = infinite, y = (x^2)/2 -(3*x)/2 + 3/2.
- Hence asymptote is a line y = (x^2)/2 - (3*x)/2 + 3/2
- Curve : x = -infinite to x = 0
- x < 0 and y < 0 : Curve is in 3rd quadrant
- x = -infinte and y = +infinte
- x = 0 and y = +infinite
- Hence there is minimum point (x1,y1)
- Curve of y is from +infinite to minimum point (x1, y1)
- Then from (x1,y1) to +infinite when x = 0
- Curve is between line y = x/2 - 1 and x = 0
- Hence curve decrease from +infinite to a minimum point (x1,y1)
- From minimum point (x1,y1) increases to infinite when x = 0
- Curve is concave upward
- Curve : x = 0 to x = +infinite
- x between 0 and 1 and y < 0 : Curve is in 4th quadrant
- x = 0 and y = -infinite
- x = 1 and y = 0 : point of inflection
- x > 1 and y increases from 0 to +infinite : Curve in 1st quadrant
- Curve of y is from -infinite to (1, 0)
- Then from (1, 0) to infinite
- Curve is between line y = x/2 - 1 and x = 0
- Hence curve increase from -infinite to a minimum point (1, 0)
- From minimum point (1,0) increases to infinite as x = infinite
- Curve is concave Downward when x is between 0 and 1
- Curve is concave upward when x is greater than 1
- From curve we see
- Minimum at (-0.5, 3.5) : Hence y' = 0 when x = -0.5
- Point of inflection at (1, 0) : Hence y" = 0 when x = 1
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Q03. Find signs of y' and y" from given curve
Signs of y'
- When x < 0
- y' > 0 if x < -0.5. Hence curve is decreasing
- y' = 0 if x = -0.5. Hence curve has minimum point (-0.5, 3.5)
- y' > 0 if x between -1 and 0. Hence curve is inreasing
- Hence the curve is between parabola asymptote and x = 0
- Hence the curve from +infinite to minimum (-0.6, 3.5)
- Then from (-0.5, 3.5) to +infinite
- Hence the curve is concave upward (y" > 0)
- When x between 0 and 1
- y' > 0 if x between 0 and 1. Curve is increasing
- y' = 0 if x = 1. Hence curve has point inflection
- Hence the curve is from -infinite to (1,0)
- Hence the curve is concave downward when x between 0 and 1. ( y"< 0)
- Hence the curve has point inflextion at x = 1. (y" = 0)
- When x > 1
- y' > 0 if x > 1. Hence curve is increasing
- Hence the curve is from (1,0) and to +infinite
- Hence the curve is concave upward when x is greater than 1. (y" > 0)
- y" > 0 if x < 0. Hence curve is concave upward
- y" < 0 if x between 0 and 1. Hence the curve is concave downward
- y" > 0 is x > 1. Hence curve is concave upward
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Q03. Find y'
Formula to find derivative of y = f(x)/g(x)
- dy/dx = (g(x)*f'(x) - g'(x)*f(x))/(g(x)^2)
- y' = (3*(x - 1)^2)*(2*x) - 2*(x - 1)^3)/(4*x^2)
- = ((x - 1)^2)*(2*x + 1))/(2*x^2)
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Q04. Find y"
Second derivative y"
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Q05. Discussion
Graphic solution
- The graphic solution clearly determine the signs of y, y' and y"
- The maximum and minimum points can only be estimated
- Without graph, the concavity can only be determined
- by the asymptotes
- By change of increasing from decreasing or vise versa
- Graphic solution will not need the knowledge of calculus
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