Find the smallest value of the polynomial x3-18x2+96x in the interval [0 , 9] .

Question

Find the smallest value of the polynomial x3-18x2+96x in the interval [0 , 9] . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let, f(x)=x3-18x2+96x Differentiating it w.r.t. x f'(x)=3x2-36x+96 =3(x2 - 12 x + 32) =3(x - 4)(x - 8) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-cfno1wan2vy7dxqh.jpg" /> Clearly at x=8f(x) is minimum. min f(x)=8(64 - 144 + 96)=128 At x=0,f(x)=0

Q. Find the smallest value of the polynomial $x^{3} - 18 x^{2} + 96 x$ in the interval $[0, 9]$ .

Let, $f (x) = x^{3} - 18 x^{2} + 96 x$
Differentiating it w.r.t. $x$
$f^{^{'}} (x) = 3 x^{2} - 36 x + 96$
$= 3 (x^{2} - 12 x + 32)$
$= 3 (x - 4) (x - 8)$

Clearly at $x = 8 f (x)$ is minimum.
$min f (x) = 8 (64 - 144 + 96) = 128$
At $x = 0, f (x) = 0$

Q. Find the smallest value of the polynomial x3−18x2+96x in the interval [0,9] .

Let, f(x)=x3−18x2+96x Differentiating it w.r.t. x f′(x)=3x2−36x+96 =3(x2−12x+32) =3(x−4)(x−8) Clearly at x=8f(x) is minimum. minf(x)=8(64−144+96)=128 At x=0,f(x)=0

Q. Find the smallest value of the polynomial $x^{3} - 18 x^{2} + 96 x$ in the interval $[0, 9]$ .

Let, $f (x) = x^{3} - 18 x^{2} + 96 x$
Differentiating it w.r.t. $x$
$f^{^{'}} (x) = 3 x^{2} - 36 x + 96$
$= 3 (x^{2} - 12 x + 32)$
$= 3 (x - 4) (x - 8)$

Clearly at $x = 8 f (x)$ is minimum.
$min f (x) = 8 (64 - 144 + 96) = 128$
At $x = 0, f (x) = 0$