To find the local maximum, local minimum, and saddle points of the function:

$f(x, y) = y^2 - 8y \cos(x)$ over the domain $-1 \leq x \leq 7$, we proceed with the following steps:

Step 1: Find the critical points

Critical points occur where the partial derivatives of $f(x, y)$ with respect to $x$ and $y$ are both zero.

1. Partial derivative with respect to $x$:

$f_x = \frac{\partial}{\partial x}(y^2 - 8y\cos(x)) = 8y \sin(x)$

Set $f_x = 0$:

$8y \sin(x) = 0$

This implies $y = 0$ or $\sin(x) = 0$. Since $\sin(x) = 0$ at $x = n\pi$ where $n$ is an integer, and we are restricted to the domain $-1 \leq x \leq 7$, we get $x = 0, \pi, 2\pi, 3\pi$, which are approximately $x = 0, 3.14, 6.28$.

2. Partial derivative with respect to $y$:

$f_y = \frac{\partial}{\partial y}(y^2 - 8y\cos(x)) = 2y - 8\cos(x)$

Set $f_y = 0$:

$2y - 8\cos(x) = 0$

$y = 4\cos(x)$

Step 2: Solve for the critical points

From the equation $y = 4\cos(x)$, substitute the values of $x = 0, \pi, 2\pi, 3\pi$:

• At $x = 0$, $y = 4\cos(0) = 4$, so the point is $(0, 4)$.
• At $x = \pi$, $y = 4\cos(\pi) = -4$, so the point is $(\pi, -4)$.
• At $x = 2\pi$, $y = 4\cos(2\pi) = 4$, so the point is $(2\pi, 4)$.
• At $x = 3\pi$, $y = 4\cos(3\pi) = -4$, so the point is $(3\pi, -4)$.

Step 3: Determine the nature of the critical points

To classify the critical points as local maxima, minima, or saddle points, we use the second partial derivative test.

1. Second partial derivatives:

   • $f_{xx} = \frac{\partial^2}{\partial x^2}(f_x) = 8y\cos(x)$
   • $f_{yy} = \frac{\partial^2}{\partial y^2}(f_y) = 2$
   • $f_{xy} = \frac{\partial^2}{\partial x \partial y}(f_x) = 8\sin(x)$

2. Discriminant (D): The discriminant is given by: $D = f_{xx}f_{yy} - (f_{xy})^2$

   Plugging in the values of the second derivatives at each critical point:

   • At $(0, 4)$: $f_{xx} = 32$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 32 \cdot 2 - 0^2 = 64 > 0$ and $f_{xx} > 0$. This point is a local minimum.

   • At $(\pi, -4)$: $f_{xx} = -32$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = (-32)(2) - 0^2 = -64 < 0$. This point is a saddle point.

   • At $(2\pi, 4)$: $f_{xx} = 32$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = 32 \cdot 2 - 0^2 = 64 > 0$ and $f_{xx} > 0$. This point is a local minimum.

   • At $(3\pi, -4)$: $f_{xx} = -32$, $f_{yy} = 2$, $f_{xy} = 0$, so $D = (-32)(2) - 0^2 = -64 < 0$. This point is a saddle point.

Step 4: Summary of results

• Local minimum values occur at $f(0, 4)$ and $f(2\pi, 4)$.
• Saddle points occur at $f(\pi, -4)$ and $f(3\pi, -4)$.

Final answers:

• Local maximum value(s): DNE
• Local minimum value(s): $f(0, 4) = -16$, $f(2\pi, 4) = -16$
• Saddle point(s): $(\pi, -4, 16)$, $(3\pi, -4, 16)$

Would you like to see a 3D graph of this function to visualize the critical points and behavior? Let me know!

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5 Related Questions:

1. How does the second derivative test classify points in two-variable functions?
2. What is the meaning of a saddle point in a function of two variables?
3. How would the results change if we altered the domain of the function?
4. Why is there no local maximum for this particular function?
5. How can one interpret the significance of cosine functions in defining the shape of a graph?

Tip: The discriminant from the second partial derivative test helps classify points as minima, maxima, or saddle points based on its sign.