2. Consider the function \( f(x, y)=e^{1+x^{2}+y^{2}} \). - Sketch, if possible, the level curves for the values \( c=-1,0,1,2 \) - Find and classify the critical points of \( f(x, y) \).

To analyze the function \( f(x, y) = e^{1 + x^2 + y^2} \), we will first look at the level curves and then find and classify the critical points. ### Level Curves The level curves of a function \( f(x, y) \) for a constant \( c \) are defined by the equation \( f(x, y) = c \). For our function, this means we need to solve: \[ e^{1 + x^2 + y^2} = c \] Taking the natural logarithm of both sides, we get: \[ 1 + x^2 + y^2 = \ln(c) \] This implies: \[ x^2 + y^2 = \ln(c) - 1 \] Now, we can analyze the values of \( c \): 1. **For \( c = -1 \)**: \[ x^2 + y^2 = \ln(-1) \] This is undefined since the logarithm of a negative number is not real. Thus, there are no level curves for \( c = -1 \). 2. **For \( c = 0 \)**: \[ x^2 + y^2 = \ln(0) \] This is also undefined since the logarithm of zero is negative infinity. Thus, there are no level curves for \( c = 0 \). 3. **For \( c = 1 \)**: \[ x^2 + y^2 = \ln(1) - 1 = 0 - 1 = -1 \] This is again undefined since \( x^2 + y^2 \) cannot be negative. Thus, there are no level curves for \( c = 1 \). 4. **For \( c = 2 \)**: \[ x^2 + y^2 = \ln(2) - 1 \] Since \( \ln(2) \) is approximately \( 0.693 \), we have: \[ x^2 + y^2 \approx 0.693 - 1 = -0.307 \] This is also undefined since \( x^2 + y^2 \) cannot be negative. Thus, there are no level curves for \( c = 2 \). ### Summary of Level Curves In summary, there are no level curves for the values \( c = -1, 0, 1, 2 \) because the logarithm of these values either leads to undefined results or negative values for \( x^2 + y^2 \). ### Critical Points To find the critical points of \( f(x, y) \), we need to compute the partial derivatives and set them to zero. 1. **Partial Derivatives**: \[ f_x = \frac{\partial f}{\partial x} = e^{1 + x^2 + y^2} \cdot 2x \] \[ f_y = \frac{\partial f}{\partial y} = e^{1 + x^2 + y^2} \cdot 2y \] 2. **Setting Partial Derivatives to Zero**: \[ f_x = 0 \implies e^{1 + x^2 + y^2} \cdot 2x = 0 \implies 2x = 0 \implies x = 0 \] \[ f_y = 0 \implies e^{1 + x^2 + y^2} \cdot 2y = 0 \implies 2y = 0 \implies y = 0 \] Thus, the only critical point is \( (0, 0) \). ### Classification of Critical Points To classify the critical point, we can use the second derivative test. We need to compute the second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = e^{1 + x^2 + y^2} (4x^2 + 2) \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = e^{1 + x^2 + y^2} (4y^2 + 2) \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = e^{1 + x^2 + y^2} (4xy) \] At the critical point \( (0, 0) \): \[ f_{xx}(