How do you calculate f''(1) and f'''(1) given f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4?

Dec 31, 2016

The Taylor series for $f \left(x\right)$ centered at $a$ is:

f(x) = f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * *

Explanation:

So the Taylor series for $f \left(x\right)$ centered at $1$ is:

f(x) = f(1)+f'(1)(x-1)+ (f''(1))/(2!)(x-1)^2+(f'''(1))/(3!)(x-1)^3+ * * *

Comparing the given function with the general expression for the Taylor series for $f$ centered at $1$, we see that

$f \left(1\right) = 3$, and $f ' \left(1\right) = - 6$, and $f ' ' \left(1\right) = 4$, and $f ' ' ' \left(1\right) = - 5$, and finallya ${f}^{\left(4\right)} \left(x\right) = 1$.

So

$f ' ' \left(1\right) = 4$, and $f ' ' ' \left(1\right) = - 5$.

If you prefer , you could differentiate to get

f(x)= 3- 6(x-1) + 4/(2!) (x-1)^2 - 5/(3!) (x-1)^3 + 1/(4!) (x-1)^4

f'(x) = -6+4(x-1)-5/(2!)(x-1)^2+1/(3!)(x-1)^3

f(x)= 4 - 5 (x-1) + 1/(2!) (x-1)^2

So

$f ' ' \left(1\right) = 4$, and $f ' ' ' \left(1\right) = - 5$.