If f(x)=x^5-5 x^4+5 x^3-10 has its local maxi | vedclass

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(A) Given $f(x) = x^5 - 5x^4 + 5x^3 - 10$. First,find the derivative $f'(x)$: $f'(x) = 5x^4 - 20x^3 + 15x^2$. For local maxima and minima,set $f'(x) =

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$5$

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(A) Given $f(x) = x^5 - 5x^4 + 5x^3 - 10$. First,find the derivative $f'(x)$: $f'(x) = 5x^4 - 20x^3 + 15x^2$. For local maxima and minima,set $f'(x) = 0$: $5x^2(x^2 - 4x + 3) = 0 \Rightarrow 5x^2(x - 1)(x - 3) = 0$. The critical points are $x = 0, 1, 3$. Now,find the second derivative $f''(x)$: $f''(x) = 20x^3 - 60x^2 + 30x$. Check the nature of the critical points: For $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 For $x = 3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$. Thus,$x = 3$ is a point of local minima. So,$b = 3$. For $x = 0$: $f''(0) = 0$,which is a point of inflection. Finally,calculate $2a + b$: $2a + b = 2(1) + 3 = 5$.

If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively,then $2 a+b$ is equal to

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