#2 - suppose that the function f(x) is approx...
#2 - suppose that the function f(x) is approximated near x = 5 by a third - degree taylor polynomial 7 - 3(x - 5)+8(x - 5)^2 - 10(x - 5)^3. find the value of f(5).
Answer
# 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 \(f^{(3)}(5)=-60\).