Question

The quadriatic approximation of $f\left(x\right)=x^3-3x^2-5$ at the point $x=0$ is

• A. $3x^2-6x-5$
• B. $-3x^2-5$
• C. $-3x^2+6x-5$
• D. $3x^2-5$

Answer 1

The correct option is B $-3x^2-5$

$f\left(x\right)=x^3-3x^2-5then$
$f^{\prime }\left(x\right)=3x^2-6x\Rightarrow f^{\prime }\left(0\right)=0$
$f^{\prime \prime }\left(x\right)=6x-6\Rightarrow f^{\prime \prime }\left(0\right)=-6$
$f\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\frac{(x-a)}{1!}+f^{\prime \prime }\left(a\right)\frac{(x-a{)}^2}{2!}+.....$
Approximation at a=0
$=f\left(0\right)+f^{\prime }\left(0\right)x+f^{\prime \prime }\left(0\right)\frac{x^2}{2!}$
$=-5+0\times x+(-6)\frac{x^2}{2}$
$=-3x^2-5$