Find the linear approximation of the function g(x) = 31 + x at a = 0.g(x)≈ Use it to approximate the numbers 30.95 and 31.1. (Round your answers to three decimal places.)30.95≈ 31.1≈

The linear approximation of a function at a certain point is given by the equation of the tangent line to the function at that point. For a function g(x), the linear approximation L(x) at a point a is given by:

L(x) = g(a) + g'(a)(x - a)

In this case, g(x) = 31 + x and a = 0. The derivative of g(x) is g'(x) = 1. So, the linear approximation at a = 0 is:

L(x) = g(0) + g'(0)(x - 0) = 31 + 1*x = 31 + x

So, the linear approximation of g(x) at a = 0 is L(x) = 31 + x.

Now, we can use this linear approximation to approximate the numbers 30.95 and 31.1.

For 30.95, we substitute x = 30.95 - 31 = -0.05 into the linear approximation:

L(-0.05) = 31 + (-0.05) = 30.95

For 31.1, we substitute x = 31.1 - 31 = 0.1 into the linear approximation:

L(0.1) = 31 + 0.1 = 31.1

So, the linear approximation of 30.95 is approximately 30.95 and the linear approximation of 31.1 is approximately 31.1.