Maximum slope of the curve y = -x^3 + 3x^2 + | vedclass

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(B) Let the function be $f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve at any point $x$ is given by the derivative $f'(x) = -3x^2 + 6x + 9$.To

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let the function be $f(x) = -x^3 + 3x^2 + 9x - 27$.The slope of the curve at any point $x$ is given by the derivative $f'(x) = -3x^2 + 6x + 9$.To find the maximum slope,let $g(x) = f'(x) = -3x^2 + 6x + 9$.We find the derivative of $g(x)$ with respect to $x$: $g'(x) = -6x + 6$.Setting $g'(x) = 0$ for critical points gives $-6x + 6 = 0$,which implies $x = 1$.To verify the maximum,we find the second derivative: $g''(x) = -6$. Since $g''(1) = -6 Substituting $x = 1$ into $g(x)$,we get the maximum slope: $g(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

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

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