Find the maxima, minima, and saddle points of the function f(x,y) = (x2 − 1)(x+2).

To find the maxima, minima, and saddle points of the function $f(x,y) = (x^2 - 1)(x+2)$, we first observe that the function only depends on $x$. We can write it as $f(x) = x^3 + 2x^2 - x - 2$.

Step 1: Find the first partial derivatives and set them to zero to find critical points.
The given function is $f(x,y) = x^3 + 2x^2 - x - 2$.
The partial derivative with respect to $x$ is:

$$f_x = \frac{\partial}{\partial x}(x^3 + 2x^2 - x - 2) = 3x^2 + 4x - 1$$

The partial derivative with respect to $y$ is:

$$f_y = \frac{\partial}{\partial y}(x^3 + 2x^2 - x - 2) = 0$$

Set $f_x = 0$ and $f_y = 0$:

1. $3x^2 + 4x - 1 = 0$
2. $0 = 0$

From equation (1), we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the values of $x$:

$$x = \frac{-4 \pm \sqrt{4^2 - 4(3)(-1)}}{2(3)}$$ $$x = \frac{-4 \pm \sqrt{16 + 12}}{6}$$ $$x = \frac{-4 \pm \sqrt{28}}{6}$$ $$x = \frac{-4 \pm 2\sqrt{7}}{6}$$ $$x = \frac{-2 \pm \sqrt{7}}{3}$$

The critical points are lines where $x = \frac{-2 + \sqrt{7}}{3}$ and $x = \frac{-2 - \sqrt{7}}{3}$.

Step 2: Find the second partial derivatives.

$$f_{xx} = \frac{\partial}{\partial x}(3x^2 + 4x - 1) = 6x + 4$$ $$f_{yy} = \frac{\partial}{\partial y}(0) = 0$$ $$f_{xy} = \frac{\partial}{\partial y}(3x^2 + 4x - 1) = 0$$

Step 3: Calculate the discriminant $D(x,y) = f_{xx}f_{yy} - (f_{xy})^2$.

$$D(x,y) = (6x + 4)(0) - (0)^2 = 0$$

Since $D(x,y) = 0$, the Second Derivative Test is inconclusive. However, since the function only depends on $x$, we can analyze the behavior of $f(x)$ using its second derivative $f''(x) = f_{xx}(x)$.

Step 4: Classify the critical points using $f_{xx}$.

• For $x_1 = \frac{-2 + \sqrt{7}}{3}$:

  $$f_{xx}\left(\frac{-2 + \sqrt{7}}{3}\right) = 6\left(\frac{-2 + \sqrt{7}}{3}\right) + 4$$ $$f_{xx}\left(\frac{-2 + \sqrt{7}}{3}\right) = 2(-2 + \sqrt{7}) + 4$$ $$f_{xx}\left(\frac{-2 + \sqrt{7}}{3}\right) = -4 + 2\sqrt{7} + 4 = 2\sqrt{7}$$

  Since $2\sqrt{7} > 0$, the critical line $x = \frac{-2 + \sqrt{7}}{3}$ corresponds to a local minimum.

• For $x_2 = \frac{-2 - \sqrt{7}}{3}$:

  $$f_{xx}\left(\frac{-2 - \sqrt{7}}{3}\right) = 6\left(\frac{-2 - \sqrt{7}}{3}\right) + 4$$ $$f_{xx}\left(\frac{-2 - \sqrt{7}}{3}\right) = 2(-2 - \sqrt{7}) + 4$$ $$f_{xx}\left(\frac{-2 - \sqrt{7}}{3}\right) = -4 - 2\sqrt{7} + 4 = -2\sqrt{7}$$

  Since $-2\sqrt{7} < 0$, the critical line $x = \frac{-2 - \sqrt{7}}{3}$ corresponds to a local maximum.

Step 5: Calculate the function values at these critical lines.
At the critical points, $3x^2 + 4x - 1 = 0$, which means $3x^2 = 1 - 4x$.
We can simplify $f(x) = x^3 + 2x^2 - x - 2$ using this relation:

$$f(x) = x(x^2) + 2x^2 - x - 2$$ $$$$