The corresponding linear approximation isx + 1 ≈ 53​ + x6        (when x is near 8).In particular, we have8.95 ≈ 53 +  6 =     (round to four decimal places)and  9.04 ≈ 53 +  6 =     (round to four decimal places).

The corresponding linear approximation isx + 6 ≈ + x6        (when x is near 3).In particular, we have8.99 ≈ 52 + 6 =     (round to four decimal places)and  9.01 ≈ 52 + 6 =     (round to four decimal places).The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when x is near 3. We also see that our approximations are overestimates because the tangent line lies above the curve.     Of course, a calculator could give us approximations for 8.99 and 9.01, but the linear approximation gives an approximation over an entire interval.

Use a linear approximation (or differentials) to estimate the given number.(1.999)3

Use a linear approximation (or differentials) to estimate the given number.100.4

Suppose that we don't have a formula for g(x) but we know that g(1) = −5 and g'(x) = x2 + 8 for all x.Use a linear approximation to estimate g(0.9) and g(1.1).g(0.9) ≈ g(1.1) ≈

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≈