y = ((x - 1)^4)/(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 3rd quadrant
- x > 0 and y > 0 : Curve in 1st quadrant
- x = 1 and y = 0
- Asymptote
- x = 0 and y = infinite. Hence x = 0 is vertical asymptote
- x = infinite, y = (x^3)/2 - 4*(x^2)/2 + 6*(x)/2 - 4/2.
- Hence asymptote is a line y = (x^3)/2 - 2*x^2 + 3*x - 2
- Curve : x = -infinite to x = 0
- x < 0 and y < 0 : Curve is in 3rd quadrant
- x = -infinite and y = -infinite
- x = 0 and y = -infinite
- Hence there is maximum point (x1, y1) in 3rd quadrant
- Curve is between line y = (x^3)/2 - 2*(x^2) + 3*x - 2 and x = 0
- Curve of y increases from -infinite to (x1, y1)
- Curve of y from (x1, y1) decreases to -infinite
- Hence curve increase from -infinite to a maximum point
- From maximum point decreases to -infinite at x = 0
- Curve is concave downward
- Curve : x = 0 to x = +infinite
- x > 0 and y > 0 : Curve is in 1st quadrant
- When x = 0 and y = +infinite
- When x = 1 and y = 0 : Minimum point
- Curve of y is from infinite to (1, 0)
- Curve is between line y = (x^3)/2 + 2*x^3 - 3*x - 2 and x = 0
- Hence curve decrease from +infinite to a minimum point
- From minimum point increases to infinite as x = infinite (1, 0)
- Curve from (1, 0) increases to infinite
- Curve is concave upward
- From curve we see
- Maximum at (-0.4,-4.7)
- Minimum at (+1, 0)
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Q03. Find signs of y' and y" from curve
Signs of y'
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Q03. Find y'
First derivative y'
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Q04. Find y"
Second derivative y"
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Q05. Discussion
Graphic solution
- The graphic solution is clear 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
- Graphic solution will not need the knowledge of calculus
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