The tangent plane of the surface \(z=f(x, y)\) at a local maximum off is parallel to the \(xy\)-plane?
TRUE
FALSE answer± The tangent plane to the function \(g(x, y, z) =z-f(x, y)\) can be written \( (-f_x,-f_y,1 \) which is \( (0,0,1) \) at critical points.
For any smooth functions \(f(x, y), x(t), y(t) \), we have \(\frac{d}{dt} f(x(t), y(t)) =f_x(x(t), y(t))x'(t) +f_y(x(t), y(t))y'(t) \)?
At a local maximum of a function \(f(x, y) \) we always have \( f_{xx} \leq 0 \) and \(f_{yy} \leq 0\).?
TRUE
FALSE answer± Indeed if one of the conditions is not satisfied, then the function increases in some direction
If a smooth function \(f(x, y) \) has a global maximum, then it has a global minimum.
TRUE
FALSE answer± A counter example is \( f(x, y) =-x^2-y^2 \).
If a point \( (x_0, y_0) \) is a minimum of \( f(x, y)\) under the constraint \( g(x, y) = 1\),then it is also a local minimum of the function \(f(x, y)\) without constraints.
TRUE
FALSE answer± The gradient of \(f\) does not have to be the zero vector..
If we extremize the function \(f(x, y)\) under the constraint \(g(x, y) = 1\), and the functions are the same \(f=g\), all points on the constraint curve are extrema for \(f\).
TRUE
FALSE answer± Every point on the curve \( g(x, y) = 1\) is a solution to the Lagrange equations because \( \nabla f=\nabla g\).
If \(f(x, y)\) is a function of two variables and (0,0) is a maximum of \(g(x) =f(x,0)\) and as well as a maximum of \(h(y) =f(0, y)\) then (0,0) is a maximum of \(f\).
TRUE
FALSE answer± \( f_{xx}>0\) and \( f_{yy}>0\) does not imply \(D=f_{xx}f_{yy}-(f_{xy})^2 \) is positive.
The vectors \(v=\langle 2,1,5\rangle \) and \(w=\langle 2,1,-1\rangle \) are perpendicular.
The tangent plane of the graph of \(f(x, y) = \sin(x)+y^3 \) at (0,1,1) is \(x+3y=3\).
TRUE
FALSE answer± In order to find the tangent plane, first write the graph as a level curve of a function of 3 variables: \(g(x, y, z) = \sin(x) +y^3-z= 0 \). Its gradient is \( \langle \cos(x),3y^2,-1 \rangle=\langle 1,3,-1 \rangle \). Therefore, tangent plane is \( x+3y-z=2\).
For any function \(f(x,y,z)\), for any unit vector \(u \) and any point \( (x_0,y_0,z_0) \) we have \(D_u f(x_0,y_0,z_0) =-D_{-u}f(x_0,y_0,z_0) \).
TRUE
FALSE answer± Changing the direction of \(u \) changes the sign by definition \(D_u f = \nabla f \cdot u \).
