The function f(x) = x^5 - 5x^4 + 5x^3 - 1 is: | vedclass

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(C) Let $f(x) = x^5 - 5x^4 + 5x^3 - 1$. First,find the derivative $f'(x)$: $f'(x) = 5x^4 - 20x^3 + 15x^2$. Set $f'(x) = 0$ to find critical points: $5

Vedclass · Accepted Answer

Neither maximum nor minimum at $x = 0$

Vedclass · Answer

(C) Let $f(x) = x^5 - 5x^4 + 5x^3 - 1$. First,find the derivative $f'(x)$: $f'(x) = 5x^4 - 20x^3 + 15x^2$. Set $f'(x) = 0$ to find critical points: $5x^2(x^2 - 4x + 3) = 0$ $5x^2(x - 3)(x - 1) = 0$. Thus,the critical points are $x = 0, 1, 3$. Now,find the second derivative $f''(x)$: $f''(x) = 20x^3 - 60x^2 + 30x$. Evaluate $f''(x)$ at the critical points: $f''(0) = 0$. $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 $f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 540 - 540 + 90 = 90 > 0$ (Local minimum at $x = 3$). For $x = 0$,since $f''(0) = 0$,we check the third derivative $f'''(x) = 60x^2 - 120x + 30$. $f'''(0) = 30 
eq 0$. Since the first non-zero derivative at $x = 0$ is of odd order ($3^{rd}$ order),$x = 0$ is a point of inflection. Therefore,the function is neither maximum nor minimum at $x = 0$.

The function $f(x) = x^5 - 5x^4 + 5x^3 - 1$ is:

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