The question is attached to this post. Please give full solution and explanation to the answer. Suppose that f (x, y, z) is a function with the following properties:f(5, 2, 10) = 8,fx (5, 2, 10) = 3, fy(5,2,10) = -2, f<(5,2, 10) = 4.%3D%3D%D%3DFind the approximate value (to three decimal places) of f (5.002, 2, 10).

Concept Used:-Linearization of a function f(x, y, z) at a point (x0, y0 ,z0) is L(x, y, z)=f(x0, y0, z0)+(x-x0)fx(x0, y0, z0)+(y-y0)fy(x0, y0, z0)+(z-z0)fz(x0, y0, z0)Given:- f(5, 2, 10)=8, fx(5, 2, 10)=3, fy(5, 2, 10)=-2, fz(5, 2, 10)=4To find:- the approximate value of f(5.002, 2, 10)Consider the linearization at the point (x0, y0, z0)=(5, 2, 10) of f(x, y, z)L(x, y, z)=f(x0, y0, z0)+(x-x0)fx(x0, y0, z0)+(y-y0)fy(x0, y0, z0)+(z-z0)fz(x0, y0, z0)Substitute the value of f(5, 2, 10)=8, fx(5, 2, 10)=3, fy(5, 2, 10)=-2, fz(5, 2, 10)=4 in the formula above we getL(x, y, z)=f(5, 2, 10)+(x-5)fx(5, 2, 10)+(y-2)fy(5, 2, 10)+(z-10)fz(5, 2, 10) =8+(x-5).3+(y-2).(-2)+(z-10).4 =8+3x-15-2y+4+4z-40 =3x-2y+4z-47Thus L(x, y, z)=3x-2y+4z-47 To find the approximate value of f(5.002, 2,10) we substitute x=5.002, y=2, z=10 in the linear approximation ConsiderL(x, y, z)=3x-2y+4z-47L(5.002, 2, 10)=3(5.002)-2(2)+4(10)-47 =15.006-4+40-47 =4.006Thus the approximate value of f(5.002, 2,10) is 4.006 (to 3 decimal places) The approximate value of f(5.002, 2,10) is 4.006 (to 3 decimal places)