The smallest value of the polynomial x3 - 18x2 + 96x in [0,9] is

Question

The smallest value of the polynomial x3 - 18x2 + 96x in [0,9] is (A) 126 (B) 0 (C) 135 (D) 160

Tardigrade Admin · Accepted Answer

f(x) = x3 - 18x2 + 96x ⇒ f'(x) = 3x2 - 36x + 96 ∴ f'(x) = 0 ⇒ x2 - 12x + 32 = 0 ⇒ x = 8, 4. Now, f(0) = 0, f(4) = 160, f(8) = 128, f(9) = 135 So, smallest value of f(x) is 0 at x = 0.

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

$126$

$0$

$135$

$160$

$f (x) = x^{3} - 18 x^{2} + 96 x$
$\Rightarrow f^{'} (x) = 3 x^{2} - 36 x + 96$
$∴ f^{'} (x) = 0$
$\Rightarrow x^{2} - 12 x + 32 = 0$
$\Rightarrow x = 8$ , $4$ .
Now, $f (0) = 0$ , $f (4) = 160$ ,
$f (8) = 128$ , $f (9) = 135$
So, smallest value of $f (x)$ is $0$ at $x = 0$ .

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

126

0

135

160

f(x)=x3−18x2+96x ⇒f′(x)=3x2−36x+96 ∴f′(x)=0 ⇒x2−12x+32=0 ⇒x=8, 4. Now, f(0)=0, f(4)=160, f(8)=128, f(9)=135 So, smallest value of f(x) is 0 at x=0.

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

$126$

$0$

$135$

$160$

$f (x) = x^{3} - 18 x^{2} + 96 x$
$\Rightarrow f^{'} (x) = 3 x^{2} - 36 x + 96$
$∴ f^{'} (x) = 0$
$\Rightarrow x^{2} - 12 x + 32 = 0$
$\Rightarrow x = 8$ , $4$ .
Now, $f (0) = 0$ , $f (4) = 160$ ,
$f (8) = 128$ , $f (9) = 135$
So, smallest value of $f (x)$ is $0$ at $x = 0$ .