The function f(x) = x^5 - 5x^4 + 5x^3 - 10 ha | vedclass

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(B) To find the local maximum,we first find the derivative of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$.$f'(x) = 5x^4 - 20x^3 + 15x^2$.Setting $f'(

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(B) To find the local maximum,we first find the derivative of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$.$f'(x) = 5x^4 - 20x^3 + 15x^2$.Setting $f'(x) = 0$ for critical points:$5x^2(x^2 - 4x + 3) = 0$.$5x^2(x - 1)(x - 3) = 0$.The critical points are $x = 0, x = 1, x = 3$.Now,we find the second derivative $f''(x) = 20x^3 - 60x^2 + 30x$.Check the sign of $f''(x)$ at the critical points:For $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10$.Since $f''(1) For $x = 3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$,so it is a local minimum.For $x = 0$: $f''(0) = 0$,and checking the sign change of $f'(x)$ around $x=0$,we see $f'(x) = 5x^2(x-1)(x-3)$,which is positive on both sides of $x=0$,so it is a point of inflection.Thus,the function has a local maximum at $x = 1$.

The function $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has a local maximum when $x$ is equal to:

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