What are the minimum and maximum values of th | vedclass

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Question

(A) Let $y = f(x) = x^5 - 5x^4 + 5x^3 - 10$.First,find the derivative: $\frac{dy}{dx} = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 3)(x - 1)

Vedclass · Accepted Answer

$-37, -9$

Vedclass · Answer

(A) Let $y = f(x) = x^5 - 5x^4 + 5x^3 - 10$.First,find the derivative: $\frac{dy}{dx} = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x - 3)(x - 1)$.Setting $\frac{dy}{dx} = 0$,we get critical points at $x = 0, 1, 3$.Now,find the second derivative: $\frac{d^2y}{dx^2} = 20x^3 - 60x^2 + 30x = 10x(2x^2 - 6x + 3)$.For $x = 0$: $\frac{d^2y}{dx^2} = 0$. Checking the third derivative: $\frac{d^3y}{dx^3} = 60x^2 - 120x + 30$. At $x = 0$,$\frac{d^3y}{dx^3} = 30 
eq 0$. Thus,$x = 0$ is a point of inflection,not a local extremum.For $x = 1$: $\frac{d^2y}{dx^2} = 10(1)(2 - 6 + 3) = -10 $f(1) = 1 - 5 + 5 - 10 = -9$.For $x = 3$: $\frac{d^2y}{dx^2} = 10(3)(2(9) - 6(3) + 3) = 30(18 - 18 + 3) = 90 > 0$. Thus,$x = 3$ is a local minimum.$f(3) = 3^5 - 5(3^4) + 5(3^3) - 10 = 243 - 405 + 135 - 10 = -37$.Therefore,the minimum value is $-37$ and the maximum value is $-9$.

What are the minimum and maximum values of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$?

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