a differentiable function ( f(x,y) ) has the property that ( f(1,2)=5 ) and ( f_x(1,2)=4 ) and ( f_y(1,2)=-7…

a differentiable function ( f(x,y) ) has the property that ( f(1,2)=5 ) and ( f_x(1,2)=4 ) and ( f_y(1,2)=-7 ).\nfind the equation of the tangent plane at the point ( (1,2,5) ) on the surface of ( z = f(x,y) ).\n( z= )

a differentiable function ( f(x,y) ) has the property that ( f(1,2)=5 ) and ( f_x(1,2)=4 ) and ( f_y(1,2)=-7 ).\nfind the equation of the tangent plane at the point ( (1,2,5) ) on the surface of ( z = f(x,y) ).\n( z= )

Answer

Explanation:

Step1: Recall the formula for the tangent plane

The formula for the tangent plane to the surface (z = f(x,y)) at the point ((x_0,y_0,z_0)) is (z=z_0 + f_x(x_0,y_0)(x - x_0)+f_y(x_0,y_0)(y - y_0))

Step2: Identify the values of (x_0,y_0,z_0,f_x(x_0,y_0),f_y(x_0,y_0))

Given ((x_0,y_0,z_0)=(1,2,5)), (f_x(1,2) = 4), (f_y(1,2)=- 7)

Step3: Substitute the values into the formula

Substitute into the formula: [ \begin{align*} z&=5+4(x - 1)+(-7)(y - 2)\ z&=5 + 4x-4-7y + 14\ z&=4x-7y+15 \end{align*} ]

Answer:

(z = 4x-7y + 15)