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  4. The maximum slope of the curve y=(1/2) x4-5 x3+18 x2-19 x occurs at the point

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

JEE MainJEE Main 2021Application of Derivatives

Solution:

$\frac{d y}{d x}=2 x^{3}-15 x^{2}+36 x-19$
Since, slope is maximum so,
$\frac{ d ^{2} y }{ dx ^{2}}=6 x ^{2}-30 x +36=0$
image
at $x=2$
$y =\frac{1}{2} \times 16-5 \times 8+18 \times 4-19 \times 2$
$=8-40+72-38=80-78=2$
point (2,2)