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

Question

The maximum slope of the curve y=(1/2) x4-5 x3+18 x2-19 x occurs at the point (A) (2,2) (B) (0,0) (C) (2,9) (D) (3, (21/2))

Tardigrade Admin · Accepted Answer

(d y/d x)=2 x3-15 x2+36 x-19 Since, slope is maximum so, ( d 2 y / dx 2)=6 x 2-30 x +36=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/cd468eea0c786789d765d630e6e92f48-.png" /> at x=2 y =(1/2) × 16-5 × 8+18 × 4-19 × 2 =8-40+72-38=80-78=2 point (2,2)

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

$(2, 2)$

$(0, 0)$

$(2, 9)$

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

$\frac{d y}{d x} = 2 x^{3} - 15 x^{2} + 36 x - 19$
Since, slope is maximum so,
$\frac{d ^{2} y}{d x ^{2}} = 6 x^{2} - 30 x + 36 = 0$

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)

Q. The maximum slope of the curve y=21​x4−5x3+18x2−19x occurs at the point

(2,2)

(0,0)

(2,9)

(3,221​)