What is the maximum slope of the curve y = -x | vedclass

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(B) Given the curve $y = f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve is given by the derivative $f'(x) = -3x^2 + 6x + 9$.Let $g(x) = f'(x) =

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$12$

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(B) Given the curve $y = f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve is given by the derivative $f'(x) = -3x^2 + 6x + 9$.Let $g(x) = f'(x) = -3x^2 + 6x + 9$ represent the slope function.To find the maximum slope,we differentiate $g(x)$ with respect to $x$:$g'(x) = -6x + 6$.Setting $g'(x) = 0$ gives $-6x + 6 = 0$,which implies $x = 1$.Now,we check the second derivative $g''(x) = -6$. Since $g''(x) The maximum slope is $g(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

What is the maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$?

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