The maximum slope of the curve y = (1)/(2) x ^ 4 - 5 x ^ 3 + 18 x ^...

\dfrac{dy}{dx} = 2{x}^{3} - 15{x}^{2} + 36x - 19$<p>Since, slope is maximum so,</p><p>$\dfrac{{d}^{2}y}{d{x}^{2}} = 6{x}^{2} - 30x + 36 = 0$</p><p>$⇒ {x}^{2} - 5x + 6 = 0$</p><p>$x = 2, 3$</p><p>at$x = 2$</p><p>$\dfrac{{d}^{3}y}{d{x}^{3}} = 12x - 30$</p><p>at$x = 2, \dfrac{{d}^{3}y}{d{x}^{3}} < 0$</p><p>So, maxima</p><p>$y = \dfrac{1}{2} \times 16 - 5 \times 8 + 18 \times 4 - 19 \times 2$</p><p>$= 8 - 40 + 72 - 38 = 80 - 78 = 2$</p><p>point \left(2, 2\right)$