The maximum slope of the curve y=12x4−5x3+18x2−19x occurs at the point

m=dydx=2x3−15x2+36x−19 dmdx=6x2−30x+36 for max. dmdx=0 x2−5x+6=0 x=2,x=3d2mdx2=12x−30<0 For x=2

The maximum slope of the curve y=12x4−5x3+18x2−19x occurs at the point

m=dydx=2x3−15x2+36x−19 dmdx=6x2−30x+36 for max. dmdx=0 x2−5x+6=0 x=2,x=3d2mdx2=12x−30<0 For x=2

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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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$(3, \frac{21}{2})$

$(2, 2)$

$(2, 9)$

$(0, 0)$

answer is B.

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$m = \frac{d y}{d x} = 2 x^{3} - 15 x^{2} + 36 x - 19 \frac{d m}{d x} = 6 x^{2} - 30 x + 36$

for max. $\frac{d m}{d x} = 0$ $x^{2} - 5 x + 6 = 0 x = 2, x = 3$

$\frac{d^{2} m}{d x^{2}} = 12 x - 30 < 0$

For $x = 2$

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The maximum slope of the curve y=12x4−5x3+18x2−19x occurs at the point

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