find point of inflection at y=X⁴-2X²

To find the point(s) of inflection for the function y = x^4 - 2x^2, we need to find the second derivative of the function, set it to zero, and check for a change in concavity. Step 1: Find the first derivative, y'. y = x^4 - 2x^2 y' = (d)/(dx)(x^4 - 2x^2) = 4x^3 - 4x Step 2: Find the second derivative, y''. y'' = (d)/(dx)(4x^3 - 4x) = 12x^2 - 4 Step 3: Set the second derivative to zero to find potential inflection points. 12x^2 - 4 = 0 12x^2 = 4 x^2 = (4)/(12) x^2 = (1)/(3) x = ± sqrt((1)/(3)) x = ± (1)/(sqrt(3)) Rationalize the denominator: x = ± sqrt(3)3 Step 4: Check for a change in concavity around these x values. We need to check the sign of y'' in intervals around x = -sqrt(3)3 and x = sqrt(3)3. • For x < -sqrt(3)3 (e.g., x=-1): y''(-1) = 12(-1)^2 - 4 = 12 - 4 = 8 > 0. (Concave up) • For -sqrt(3)3 < x < sqrt(3)3 (e.g., x=0): y''(0) = 12(0)^2 - 4 = -4 < 0. (Concave down) • For x > sqrt(3)3 (e.g., x=1): y''(1) = 12(1)^2 - 4 = 12 - 4 = 8 > 0. (Concave up) Since the concavity changes at both x = -sqrt(3)3 and x = sqrt(3)3, these are indeed the x-coordinates of the points of inflection. Step 5: Find the corresponding y-coordinates by substituting the x values into the original function y = x^4 - 2x^2. For x = sqrt(3)3: y = (sqrt(3)3)^4 - 2(sqrt(3)3)^2 y = ((3)/(9))^2 - 2((3)/(9)) y = ((1)/(3))^2 - 2((1)/(3)) y = (1)/(9) - (2)/(3) y = (1)/(9) - (6)/(9) y = -(5)/(9) So, one point of inflection is (sqrt(3)3, -(5)/(9)). For x = -sqrt(3)3: Since the powers in the original function (x^4 and x^2) are even, y will be the same for x = -sqrt(3)3 as for x = sqrt(3)3. y = (-sqrt(3)3)^4 - 2(-sqrt(3)3)^2 = (1)/(9) - (6)/(9) = -(5)/(9) So, the other point of inflection is (-sqrt(3)3, -(5)/(9)). The points of inflection are (sqrt(3)3, -(5)/(9)) and (-sqrt(3)3, -(5)/(9)). What's next?