The maximum slope of the curve y = frac{1}{2} | vedclass

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(A) Let the slope of the curve be $m(x) = \frac{dy}{dx}$.$m(x) = \frac{d}{dx} \left( \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x \right) = 2x^{3} -

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$(2, 2)$

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(A) Let the slope of the curve be $m(x) = \frac{dy}{dx}$.$m(x) = \frac{d}{dx} \left( \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x \right) = 2x^{3} - 15x^{2} + 36x - 19$.To find the maximum slope,we find the critical points of $m(x)$ by setting its derivative to zero:$m'(x) = \frac{d^{2}y}{dx^{2}} = 6x^{2} - 30x + 36 = 0$.Dividing by $6$,we get $x^{2} - 5x + 6 = 0$,which factors as $(x-2)(x-3) = 0$.Thus,the critical points are $x = 2$ and $x = 3$.To check for the maximum,we use the second derivative test on $m(x)$:$m''(x) = \frac{d^{3}y}{dx^{3}} = 12x - 30$.At $x = 2$,$m''(2) = 12(2) - 30 = 24 - 30 = -6 At $x = 3$,$m''(3) = 12(3) - 30 = 36 - 30 = 6 > 0$. Since the second derivative is positive,$m(x)$ has a local minimum at $x = 3$.Therefore,the maximum slope occurs at $x = 2$.Substituting $x = 2$ into the original equation $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$:$y = \frac{1}{2}(16) - 5(8) + 18(4) - 19(2) = 8 - 40 + 72 - 38 = 2$.The point is $(2, 2)$.

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point

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