Multivariable Linear Approximation – A multiplication with integer powers in 3 variables – Exercise 4223


Find an approximate value of

1.002\cdot 2.003^2\cdot 3.004^3

Final Answer



We will use the 3-variable linear approximation formula

f(x,y,z)\approx f(x_0,y_0,z_0)+f'_x(x_0,y_0,z_0)\cdot(x-x_0)+


Therefore, we need to define the following

x, y, z, x_0, y_0, z_0, f(x,y,z)

And place them in the formula. x, y and z will be the numbers that appear in the question, and the points


will be the closest values to x, y and z respectively, which are easy for calculations.

In our exercise, we will define

x=1.002, y=2.003,z=3.004

because these values appear in the question. And we set

x_0=1, y_0=2,z_0=3

because they are the closest values to x, y, and z respectively that are easy for calculations.

After we have defined x, y and z, it is easy to find the function. Just put x instead of the number we set to be x, put y instead of the number we set to be y and put z instead of the number we set to be z. In this way, we get the function


In the formula above we see the function partial derivatives. Hence, we calculate them.




We put all the data in the formula and get

f(1.002,2.003,3.004)\approx f(1,2,3)+f'_x(1,2,3)\cdot(1.002-1)+


= 1\cdot 2^2\cdot 3^3+2^2\cdot 3^3\cdot 0.002+2\cdot 1\cdot 2\cdot 3^3\cdot 0.003+3\cdot 1\cdot 2^2\cdot 3^2\cdot 0.004=

= 108+108\cdot 0.002+108\cdot 0.003+108\cdot 0.004=

= 108(1+0.002+0.003+0.004)=

= 108\cdot 1.009=

= 108.972

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