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At $\mathrm{x}=0, f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$
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The correct answer is:
has no extremum value
We are given that $f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$
$\begin{aligned} & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\sin \mathrm{x}+\mathrm{x}-\mathrm{x}^2 \\ & \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\cos \mathrm{x}+1-2 \mathrm{x}\end{aligned}$
At $x=0$
$\mathrm{f}^{\prime \prime}(0)=-\cos 0+1-0=-1+1=0$
$\Rightarrow ' x=0$ ' is neither maxima nor minima point
$\Rightarrow A t{ }^{\prime} \mathrm{x}=0$ ' $\mathrm{f}(\mathrm{x})$ has no extremum value.
$\begin{aligned} & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\sin \mathrm{x}+\mathrm{x}-\mathrm{x}^2 \\ & \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\cos \mathrm{x}+1-2 \mathrm{x}\end{aligned}$
At $x=0$
$\mathrm{f}^{\prime \prime}(0)=-\cos 0+1-0=-1+1=0$
$\Rightarrow ' x=0$ ' is neither maxima nor minima point
$\Rightarrow A t{ }^{\prime} \mathrm{x}=0$ ' $\mathrm{f}(\mathrm{x})$ has no extremum value.
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