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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
Find the local maximum and local minimum value of $f(x) = x^3 − 3x^2 − 24x + 5$
✓

Answer

$f(x) = x^3 − 3x^2 − 24x + 5$
$\therefore f′(x) = 3x^2 – 6x – 24$
$\therefore f''(x) = 6x − 6$
Consider, $f′(x) = 0$
$\therefore 3x^2 – 6x – 24 = 0$
$\therefore 3(x^2 – 2x – 8) = 0$
$\therefore x^2 – 2x – 8 = 0$
$\therefore (x + 2)(x – 4) = 0$
$\therefore x + 2 = 0$ or $x – 4 = 0$
$\therefore x = – 2$ or $x = 4$
For $x = – 2,$
$f ''(– 2) = 6(– 2) − 6$
$= −18 < 0$
$\therefore f(x)$ is maximum at $x = – 2.$
$\therefore $ Maximum value $= f(–2) = (–2)^3 − 3(–2)^2 – 24(–2) + 5$
$= – 8 – 12 + 48 + 5$
$= 33$
For $x = 4,$
$f''(4) = 6(4) − 6$
$= 18 > 0$
$\therefore f(x)$ is minimum at $x = 4.$
$\therefore $ Minimum value $= f(4) = (4)^3 − 3 (4)^2 – 24 (4) + 5$
$= 64 − 48 – 96 + 5$
$= −75$
$\therefore $ Local maximum of $f(x)$ is $33$ when $x = – 2$ and Local minimum of $f(x)$ is $−75$ when $x = 4.$

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