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The maximum and minimum values of $x^3-18 x^2+96 x$ in interval $(0,9)$ are
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Verified Answer
The correct answer is:
160
Let $y=x^3-18 x^2+96 x \Rightarrow \frac{d y}{d x}=3 x^2-36 x+96=0$
$\therefore \quad x^2-12 x+32=0 \Rightarrow(x-4)(x-8)=0 \quad, \quad x=4,8$
Now, $\quad \frac{d^2 y}{d x^2}=6 x-36$
At $x=4 \frac{d^2 y}{d x^2}=24-36=-12 \lt 0$
$\therefore$ At $x=4$ function will be maximum
and $[f(4)]_{\text {max. }}=64-288+384=160$
At $x=8, \quad \frac{d^2 y}{d x^2}=48-36=12\gt0$
$\therefore$ At $x=8$, function will be minimum and $[f(8)]_{\text {min. }}=128$.
$\therefore \quad x^2-12 x+32=0 \Rightarrow(x-4)(x-8)=0 \quad, \quad x=4,8$
Now, $\quad \frac{d^2 y}{d x^2}=6 x-36$
At $x=4 \frac{d^2 y}{d x^2}=24-36=-12 \lt 0$
$\therefore$ At $x=4$ function will be maximum
and $[f(4)]_{\text {max. }}=64-288+384=160$
At $x=8, \quad \frac{d^2 y}{d x^2}=48-36=12\gt0$
$\therefore$ At $x=8$, function will be minimum and $[f(8)]_{\text {min. }}=128$.
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