Question

(2.001)^5 use linear approximations

Answer 1

32.08

Answer 2

To find linear approximations, you use the formula\[f(x)\approx f(a)+f'(a)(x-a)\]for some nice value of a.

Your function f(x), and it's derivative are \[f(x)=x^{5}\]\[f'(x)=5x^{4}\]
Since you are looking for f(2.001), x=2.001. By setting a=2 (a nice number for x^5), you get

\[f(x)=f(2.001)\approx f(2)+f'(2)(2.001-2)=2^{5}+(5)(2)^{4}(.001)\]\[=32+(5)(16)(.001)=32+80(.001)=32+.08=32.08\]