I think this the correct solution. but please correct me if I'm wrong
Answer: equation of the tangent plane is z = 1
Step-by-step explanation:
Given equation
z = e^(-x²-y²) at point (0,0,1)
now let z = f(x,y)
Δf(x,y) = [ fx, fy ]
= (-2xe^(-x²-y²)), (-2ye^(-x²-y²))
now
Δf (0,0) = [ 0, 0 ] = [ a, b ]
equation of the tangent plane therefore will be
z - z₀ = a(x-x₀) + b(y-y₀)
z - 1 = 0(x-0) + 0(y-0)
z - 1 = 0 + 0
z = 1
Therefore equation of the tangent plane is z = 1
g(-2) + g(2) = 22
Step-by-step explanation:
For x = -2, g(x) = x^2 - 3x
or
g(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10
For x = 2, g(x) = 12, therefore
g(-2) + g(2) = 10 + 12 = 22
Answer:
24
Step-by-step explanation:
you can just search it 20% of 120