At x=0,f(x)=cos x-1+frac{x^2}{2}-frac{x^3}{3} | vedclass

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(C) Given $f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$.First,find the first derivative $f'(x)$:$f'(x) = -\sin x + x - x^2$.Next,find the second derivat

Vedclass · Accepted Answer

has no extremum value

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(C) Given $f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$.First,find the first derivative $f'(x)$:$f'(x) = -\sin x + x - x^2$.Next,find the second derivative $f''(x)$:$f''(x) = -\cos x + 1 - 2x$.Evaluate at $x=0$:$f'(0) = -\sin(0) + 0 - 0^2 = 0$.$f''(0) = -\cos(0) + 1 - 2(0) = -1 + 1 = 0$.Since $f'(0)=0$ and $f''(0)=0$,we check the third derivative $f'''(x)$:$f'''(x) = \sin x - 2$.Evaluate at $x=0$:$f'''(0) = \sin(0) - 2 = -2$.Since the first non-zero derivative at $x=0$ is of odd order (the third derivative),$x=0$ is a point of inflection and not an extremum point.Therefore,$f(x)$ has no extremum value at $x=0$.

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

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