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, 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)-2/3^3, f(-3)=(7)6/3^4, and f^(4)(-3)=(7)-24/3^5. submit skip (you cannot come back) need help? read it submit answer
Answer
Explanation:
Step1: Recall derivative rules
The power - rule for differentiation states that if (y = \frac{k}{x^n}=kx^{-n}), then (y^\prime=-nkx^{-(n + 1)}). For (f(x)=\frac{7}{x}=7x^{-1}), (f^\prime(x)=(7)(- 1)x^{-2}), (f^{\prime\prime}(x)=(7)(-1)(-2)x^{-3}), (f^{\prime\prime\prime}(x)=(7)(-1)(-2)(-3)x^{-4}), (f^{(4)}(x)=(7)(-1)(-2)(-3)(-4)x^{-5}).
Step2: Evaluate at (a=-3)
When (a = - 3):
- (f(-3)=\frac{7}{-3}=-\frac{7}{3}).
- (f^\prime(-3)=(7)(-1)\frac{1}{(-3)^2}=-\frac{7}{9}).
- (f^{\prime\prime}(-3)=(7)(-1)(-2)\frac{1}{(-3)^3}=\frac{14}{-27}=-\frac{14}{27}).
- (f^{\prime\prime\prime}(-3)=(7)(-1)(-2)(-3)\frac{1}{(-3)^4}=-\frac{42}{81}=-\frac{14}{27}).
- (f^{(4)}(-3)=(7)(-1)(-2)(-3)(-4)\frac{1}{(-3)^5}=\frac{168}{-243}=-\frac{56}{81}).
So the blanks should be filled as follows:
- (f(-3)=(7)\left(-\frac{1}{3}\right))
- (f^\prime(-3)=(7)\left(-\frac{1}{3^2}\right))
- (f^{\prime\prime}(-3)=(7)\left(\frac{2}{3^3}\right))
- (f^{\prime\prime\prime}(-3)=(7)\left(-\frac{6}{3^4}\right))
- (f^{(4)}(-3)=(7)\left(\frac{24}{3^5}\right))
Answer:
(-\frac{1}{3},-\frac{1}{3^2},\frac{2}{3^3},-\frac{6}{3^4},\frac{24}{3^5})