Question:
The maximum slope of the curve $y=\frac{1}{2} x^{4}-5 x^{3}+18 x^{2}-19 x$ occurs at the point
Correct Option: 1
Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}^{3}-15 \mathrm{x}^{2}+36 \mathrm{x}-19$
Since, slope is maximum so,
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \mathrm{x}^{2}-30 \mathrm{x}+36=0$
$y=\frac{1}{2} \times 16-5 \times 8+18 \times 4-19 \times 2$
$=8-40+72-38=80-78=2$
point $(2,2)$
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