If x = p and x = q are the points of local ma | vedclass

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(C) Let $f(x) = x^5 - 5x^4 + 5x^3 - 10$.First,find the derivative: $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$.For critic

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$p = 1, q = 3$

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(C) Let $f(x) = x^5 - 5x^4 + 5x^3 - 10$.First,find the derivative: $f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 1)(x - 3)$.For critical points,set $f'(x) = 0$,which gives $x = 0, 1, 3$.Now,find the second derivative: $f''(x) = 20x^3 - 60x^2 + 30x$.Check the nature of critical points:At $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 At $x = 3$: $f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 540 - 540 + 90 = 90 > 0$. Since $f''(3) > 0$,$x = 3$ is a point of local minima. Thus,$q = 3$.At $x = 0$: $f''(0) = 0$. Using the first derivative test,$f'(x)$ does not change sign around $x = 0$ (it is positive on both sides),so $x = 0$ is a point of inflection,not a local extremum.Therefore,$p = 1$ and $q = 3$.

If $x = p$ and $x = q$ are the points of local maxima and local minima respectively for the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$,then:

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