joe g asked in Science & MathematicsMathematics · 10 months ago

# Help Solve?

The function y=x^4-18x^2+40 has the following points of inflection:

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• 10 months ago

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 x-values. Plug those into the original function to find the corresponding y-values.

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)

• 10 months ago

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.