Yahoo Answers: Answers and Comments for Help Solve? [Mathematics]
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From joe g
enAU
Wed, 20 Nov 2019 05:19:54 +0000
3
Yahoo Answers: Answers and Comments for Help Solve? [Mathematics]
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https://au.answers.yahoo.com/question/index?qid=20191120051954AA0JvQm
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From Puzzling: Original function:
y = x⁴  18x² + 40
Points ...
https://au.answers.yahoo.com/question/index?qid=20191120051954AA0JvQm
https://au.answers.yahoo.com/question/index?qid=20191120051954AA0JvQm
Wed, 20 Nov 2019 13:46:21 +0000
Original function:
y = x⁴  18x² + 40
Points of inflection are where the function is changing from concave up to concave down, or vice versa. At that point, the second derivative is zero.
So figure out the second derivative:
y' = 4x³  36x
y'' = 12x²  36
Set that to zero:
12x²  36 = 0
12x² = 36
x² = 3
x = ±√3
UPDATE:
To be thorough, you should verify these are actual points of inflection.
Test a point in the interval (∞, √3)
y" = 12(2)²  36 = 48  36 = 12 (positive = concave up)
Test a point in the interval (√3, √3)
y" = 12(0)²  36 = 0  36 = 36 (negative = concave down)
Test a point in the interval (√3, ∞)
y" = 12(2)²  36 = 48  36 = 12 (positive = concave up)
So those two points do occur where the concavity changes and are points of inflection. Those are the two xvalues. Plug those into the original function to find the corresponding yvalues.
y = (√3)⁴  18(√3)² + 40
y = 9  54 + 40
y = 5
(√3, 5)
y = (√3)⁴  18(√3)² + 40
y = 9  54 + 40
y = 5
(√3, 5)

From Captain Matticus, LandPiratesInc: y = x^4  18x^2 + 40
y' = 4x^3  36x
y'...
https://au.answers.yahoo.com/question/index?qid=20191120051954AA0JvQm
https://au.answers.yahoo.com/question/index?qid=20191120051954AA0JvQm
Wed, 20 Nov 2019 05:22:50 +0000
y = x^4  18x^2 + 40
y' = 4x^3  36x
y'' = 12x^2  36
y'' = 12 * (x^2  6)
y'' = 0
0 = 12 * (x^2  6)
0 = x^2  6
6 = x^2
x = +/ sqrt(6)
y = x^2 * (x^2  18) + 40
y = 6 * (6  18) + 40
y = 6 * (12) + 40
y = 72 + 40
y = 32
(sqrt(6) , 32) , (sqrt(6) , 32)
Those are the points of inflection.