Linearization(calc 3)

L(x,y) of the function f(x,y) = x^2y^3 - 4sin(x+2y) at the point P = (-2,1). Also, use it to evaluate
f(-2.1,1.05). Any help would be much appreciated

First, we find the equation of the tangent plane to f at (-2, 1).

f(-2, 1) = 4

f_x = 2xy^3 - 4 cos(x+2y) ==> f_x (-2, 1) = -8
f_y = 3x^2 y^2 - 8 cos(x+2y) ==> f_y (-2, 1) = 4.

So, z = 4 + -8(x + 2) + 4(y - 1)

Hence, the linearization L is
L(x, y) = 4 - 8(x + 2) + 4(y - 1).

Finally, since (-2.1, 1.05) is sufficiently close to (-2, 1),
f(-2.1, 1.05) ≈ L(-2.1, 1.05) = 4 - 8(-0.1) + 4(0.5) = 6.8.

I hope this helps!

fx=2xy^3-4cos(x+2y)=>
fx(-2,1)=-8
fy=3x^2y^2-8cos(x+2y)=>
fy(-2,1)=4
f(-2,1)=4
=>
L(-2.1,1.05)=4+(-0.1)(-8)+(0.05)(4)=>
L(-2.1,1.05)=4+0.8+0.2=5