Question

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 ).
find the equation of the tangent plane at the point ( (1,2,5) ) on the surface of ( z = f(x,y) ).
( z= )

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:

$$
LATEXBLOCK0
$$

Answer:

\(z = 4x-7y + 15\)