Answer to Question #216140 in Calculus for Isha

Question

f(x) =10x^3-3x^2-5x-1

Answer 1

Maclaurin series of function $f(x)$ is defined as

$$f(x)=f(0)+\dfrac{f'(0)}{1!}x+\dfrac{f''(0)}{2!}x^2+\dfrac{f'''(0)}{3!}x^3+\dfrac{f''''(0)}{4!}x^4+...$$

$f(x) =10x^3-3x^2-5x-1$

$f(0)=-1$

$f'(x)=30x^2-6x-5, f'(0)=-5$

$f''(x)=60x-6, f''(0)=-6$

$f'''(x)=60, f'''(0)=60$

$f^{(IV)}(x)=f^{(V)}(x)=f^{(VI)}(x)=...=0$

Then

$$f(x)=-1+\dfrac{-5}{1}x+\dfrac{-6}{2}x^2+\dfrac{60}{6}x^3$$

$$c_0=-1, c_1=-5, c_2=-3, c_4=10, c_5=c_6=...=0$$

$$f(x)=-1-5x-3x^2+10x^3$$

The radius of convergence $R=INF.$