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

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"Given, $y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$

${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$

$\Rightarrow$ Slope M = $2{x^3} - 15{x^2} + 36x - 19$

At max of slope ${{dM} \over {dx}} = 0$

$\therefore$ ${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$

$\Rightarrow 6({x^2} - 5x + 6) = 0$

$\Rightarrow 6(x - 2)(x - 3) = 0$

$\therefore$ $x = 2,3$

Now, ${{{d^2}M} \over {d{x^2}}} = 6(2x - 5)$

at $x = 2,{{{d^2}M} \over {d{x^2}}} = 6(4 - 5) = - 6 < 0$

$\therefore$ at x = 2 slope is maximum.

At x = 2,

y = 8 - 40 + 72 - 38 = 2

$\therefore$ Required point = (2, 2)"

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

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

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