Question

At $x=0$, $f\left(x\right)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{6}$

• A. Has a minimum
• B. Has a maximum
• C. Does not have an extremum
• D. It is not defined

Answer 1

Answer: (C)

Does not have an extremum

x=0; $f\left(x\right)=cosx-1+\frac{x^2}{2}-\frac{x^3}{6}$

$f^{\prime }\left(x\right)=sinx+\frac{2x}{2}-\frac{3x^2}{6}=0$

$f^{\prime }\left(x\right)=0\to$ at x=0

$f^{\prime \prime }\left(x\right)=-cosx+1-2$

$f^{\prime \prime }\left(0\right)=-1+1-0=0$

$\therefore$ it is a Rt. of in Hexaion

$\therefore$ not have an extremum