Question

Find all local maximum and minimum points by the second derivative test when possible y=x^4-2×^2+3​

Answer 1

y = x^4 - 2x^2 + 3​

y' = 4x^3 - 4x

y'' = 12x^2 - 4

To find local minimum and local maximum, we find solution of y' = 0

=> 4x^3 - 4x = 0

<=>4x(x - 1)(x + 1) = 0

<=> x = 0,

substitute into y => y(0) = 0 - 0 + 3 = 3 => A(0, 3)

substitute into y'' => y''(0) = 0 - 4 = -4 < 0 => A(0, 3) is a maximal

<=> x = 1,

substitute into y => y(1) = 1 - 2 + 3 = 2 => B(1, 2)

substitute into y'' => y''(1) = 12 - 4 = 8 > 0 => B(1, 2) is a minimal

<=> x = -1,

substitute into y => y(-1) = 1 - 2 + 3 = 2 => C(-1, 2)

substitute into y'' => y''(-1) = 12 - 4 = 8 > 0 => C(-1, 2) is a minimal

Hope this helps!

:)