The least value of the function f(x)=x^3-18 x^2+96 x in the inter… - Vidyadip

MCQ

The least value of the function $f(x)=x^3-18 x^2+96 x$ in the interval $[0,9]$ is:

A
$126$
B
$135$
C
$160$
✓
$0$

✓

Answer

Correct option: D.

$0$

Given$, f(x)=x^3-18 x^2+96 x$
Implies that $f^{\prime}(x)=3 x^2-36 x+96$
For a local maxima or a local minima, we must have $f^{\prime}(x)=0$
Implies that $3 x^2-36 x+96$
Implies that $x^2-12 x+32=0$
Implies that $(x-4)(x-8)=0$
Implies that $x=4,8$
Therefore, $f(8)=(8)^3-18(8)^2+96(8)=512-1152+768=128$
$f(4)=(4)^3-18(4)^2+96(4)=64-288+384=160$
$f(0)=(0)^3-18(0)^2+96(0)=0$
$f(9)=(9)^3-18(9)^2+96(9)=729-1458+864=135$
Hence, $0$ is the minimum value in the range $0,9 .$

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