Tangent Planes and Linear Approximations — Question 4

Question 4 – Solution

Strategy. A tangent plane to a graph is horizontal precisely when both of its slope coefficients vanish.

Step 1: Slope conditions $$f_x=3x^2-3,\qquad f_y=2y-4.$$ Thus a horizontal plane requires $$3x^2-3=0,\qquad 2y-4=0.$$ Therefore $x=\pm 1$ and $y=2$.

Step 2: Heights $$f(1,2)=1-3+4-8=-6,$$ $$f(-1,2)=-1+3+4-8=-2.$$ The two points are $$\boxed{(1,2,-6)\quad\text{and}\quad(-1,2,-2)}.$$

Step 3: Planes and completeness Because both slope coefficients are zero, the planes are $$\boxed{z=-6}\qquad\text{and}\qquad\boxed{z=-2}.$$ The equations $x^2=1$ and $y=2$ exhaust every simultaneous zero of $f_x$ and $f_y$, so there are no other horizontal tangent planes.