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

### 4 Answers

- PopeLv 71 month agoFavourite answer
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)

- PhilipLv 61 month ago
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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- ted sLv 71 month ago
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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- 1 month ago
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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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.