#2 - suppose that the function f(x) is approximated near x = 5 by a third - degree taylor polynomial 7 - 3(x…

Answer:

-60

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function (f(x)) about (x = a) is (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}+\cdots). For a third - degree Taylor polynomial about (x = 5), (P_3(x)=f(5)+f^{\prime}(5)(x - 5)+\frac{f^{\prime\prime}(5)}{2}(x - 5)^{2}+\frac{f^{(3)}(5)}{6}(x - 5)^{3}).

Step2: Identify the coefficient of ((x - 5)^{3})

We are given (P_3(x)=7-3(x - 5)+8(x - 5)^{2}-10(x - 5)^{3}). The coefficient of ((x - 5)^{3}) in the Taylor polynomial is (\frac{f^{(3)}(5)}{6}).

Step3: Solve for (f^{(3)}(5))

Set (\frac{f^{(3)}(5)}{6}=- 10). Then, multiply both sides of the equation by (6) to get (f^{(3)}(5)=-60).