A function of two variables is given by,

f(x,y) = e^2x-3y

Find the tangent approximation to f(0.989,1.166) near (0,0), giving your answer to 4 decimal places.

To find the tangent approximation to f(0.989,1.166) near (0,0), we can use the concept of linearization. The linearization of a multi-variable function is given by the equation:

L(x,y) = f(a,b) + (∂f/∂x)(a,b)(x-a) + (∂f/∂y)(a,b)(y-b)

where (a,b) is the point at which we want to approximate the function and (∂f/∂x)(a,b) and (∂f/∂y)(a,b) are the partial derivatives of f with respect to x and y, evaluated at (a,b).

In our case, the given function is f(x,y) = e^(2x-3y). So, let's calculate the partial derivatives.

∂f/∂x = (∂/∂x)(e^(2x-3y)) = 2e^(2x-3y)
∂f/∂y = (∂/∂y)(e^(2x-3y)) = -3e^(2x-3y)

Now, let's evaluate the partial derivatives (∂f/∂x)(0,0) and (∂f/∂y)(0,0) at (a,b) = (0,0):

(∂f/∂x)(0,0) = 2e^(2(0)-3(0)) = 2e^0 = 2
(∂f/∂y)(0,0) = -3e^(2(0)-3(0)) = -3e^0 = -3

Next, let's plug these values into the linearization formula. We have:

L(x,y) = f(0,0) + (∂f/∂x)(0,0)(x-0) + (∂f/∂y)(0,0)(y-0)

To find f(0,0), substitute x = 0 and y = 0 into the given function:

f(0,0) = e^(2(0)-3(0)) = e^0 = 1

Hence, the linearization becomes:

L(x,y) = 1 + 2x - 3y

Finally, substitute the values of x = 0.989 and y = 1.166 into the linearization equation to find the tangent approximation:

L(0.989, 1.166) = 1 + 2(0.989) - 3(1.166) = 1 + 1.978 - 3.498 = -0.52

Therefore, the tangent approximation to f(0.989,1.166) near (0,0) is -0.52, rounded to 4 decimal places.