Set the derivative equal to zero.
4 x^3 + 6x^2 − 4x = 02 x(2x^2 + 3x − 2) = 02 x(2x − 1)(x + 2) = 0x = 0, , − 2Next, plug these three values into the original equation to find the y-coordinates of the critical points. We
already know that when x = 0, y = 1.
When x = , y = = When x = −2, y = (−2)^4 + 2(−2)^3 − 2(−2)^2 + 1 = −7Thus, we have critical points at (0, 1), , and (−2, −7).
Step 3: Take the second derivative to find any points of inflection.
= 12x^2 + 12x −4Set this equal to zero.
12 x^2 + 12x − 4 = 03 x^2 + 3x − 1 = 0x = ≈ .26, − 1.26Therefore, the curve has points of inflection at x = .
Now solve for the y-coordinates.
(0.26, 0.90) and (−1.26, −3.66)We can now plug the critical values into the second derivative to determine whether each is a maximum