Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# In the given curve, compute the location of the point of inflection, y=3x + (x+2)^3/5?

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I see some people telling you that there is no point of inflection. We see so much confusion over the order of operations. Now that someone is doing it correctly, it is difficult to see. Here y is equated to a cubic polynomial. All curves of that kind have one point of inflection.

y = 3x + (x + 2)³/5

dy/dx = 3 + 3(x + 2)²/5

d²y/dx² = 6(x + 2)/5

Let d²y/dx² = 0.

6(x + 2)/5 = 0

x = -2

y = 3(-2) + [(-2) + 2]³/5

y = -6

point of inflection: (-2, -6)

• It is neither. The first derivative has degree 2. In the give equation you must have taken 3/5 as an exponent. It is not. Have a care with the order of operations. The 5 is not part of the exponent.

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• Point of inflection occurs where y'' = 0. y = 3x + (x+2)^(3/5), y' = 3+ (3/5)(x+2)^(-2/5),

y'' = (3/5)(-2/5)(x+2)^(-7/5). At no point does y'' = 0. No point of inflection exists on y.

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• Here you need to be careful about how YOU have defined "point of inflection " ...for this y function the concavity changes at - 2 so ( -2 , -6 ) is a point of inflection.......setting y ' ' = 0 yields CANDIDATES for points of inflection....eg y = x³ has y ' ' = 6x & x = 0 as a point  of inflection   while y = x^4 has y ' ' =0 but x = 0 is not a point of inflection...concavity is the important criteria

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• y = 3x + (x + 2)^(3/5)

y' = 3 + (3/5) * (x + 2)^(-2/5)

y'' = (3/5) * (-2/5) * (x + 2)^(-7/5)

y'' = 0

0 = (-6/25) * (x + 2)^(-7/5)

0 = (x + 2)^(-7/5)

No solution.

There is no inflection point.

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