step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 +…

step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+... the function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7)-1/x^2, f(x)=(7)2/x^3, f(x)=(7)-6/x^4, and f^(4)(x)=(7)24/x^5. step 2 with a = -3, f(-3)=(7)-1/3, f(-3)=(7)-1/3^2, f(-3)=(7)x, f(-3)=(7)-6/3^4, and f^(4)(-3)=(7)x. submit skip (you cannot come back)

step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+... the function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7)-1/x^2, f(x)=(7)2/x^3, f(x)=(7)-6/x^4, and f^(4)(x)=(7)24/x^5. step 2 with a = -3, f(-3)=(7)-1/3, f(-3)=(7)-1/3^2, f(-3)=(7)x, f(-3)=(7)-6/3^4, and f^(4)(-3)=(7)x. submit skip (you cannot come back)

Answer

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of a function (f(x)) about (x = a) is (f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{\prime\prime\prime}(a)}{3!}(x - a)^{3}+\frac{f^{(4)}(a)}{4!}(x - a)^{4}+\cdots)

Step2: Find function and its derivatives

Given (f(x)=\frac{7}{x}=7x^{-1}). Using the power - rule for differentiation ((x^n)^\prime=nx^{n - 1}), we have:

  • (f^{\prime}(x)=7\times(-1)x^{-2}=\frac{-7}{x^{2}})
  • (f^{\prime\prime}(x)=7\times(-1)\times(-2)x^{-3}=\frac{14}{x^{3}})
  • (f^{\prime\prime\prime}(x)=7\times(-1)\times(-2)\times(-3)x^{-4}=\frac{-42}{x^{4}})
  • (f^{(4)}(x)=7\times(-1)\times(-2)\times(-3)\times(-4)x^{-5}=\frac{168}{x^{5}})

Step3: Evaluate at (a=-3)

  • (f(-3)=\frac{7}{-3}=-\frac{7}{3})
  • (f^{\prime}(-3)=\frac{-7}{(-3)^{2}}=-\frac{7}{9})
  • (f^{\prime\prime}(-3)=\frac{14}{(-3)^{3}}=-\frac{14}{27})
  • (f^{\prime\prime\prime}(-3)=\frac{-42}{(-3)^{4}}=-\frac{42}{81}=-\frac{14}{27})
  • (f^{(4)}(-3)=\frac{168}{(-3)^{5}}=-\frac{168}{243}=-\frac{56}{81})

The values in step 2 for (f^{\prime\prime}(-3)) and (f^{(4)}(-3)) in the original were incorrect. The correct values are (f^{\prime\prime}(-3)=(7)\frac{2}{(-3)^{3}}) and (f^{(4)}(-3)=(7)\frac{24}{(-3)^{5}})

Answer:

The correct values for (f^{\prime\prime}(-3)) and (f^{(4)}(-3)) in step 2 are (f^{\prime\prime}(-3)=(7)\frac{2}{(-3)^{3}}) and (f^{(4)}(-3)=(7)\frac{24}{(-3)^{5}})