Determine the maximum and minimum value of the following function… - Vidyadip

Question

Determine the maximum and minimum value of the following function.$f(x) = 2x^3 – 21x^2 + 36x – 20$

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Answer

$f(x) = 2x^3 – 21x^2 + 36x – 20\therefore f'(x) = 6x^2 – 42x + 36$ and$ f''(x) = 12x – 42$
Consider,$ f '(x) = 0$
$\therefore 6x^2 – 42x + 36 = 0$
$\therefore 6(x^2 – 7x + 6) = 0$
$\therefore 6(x – 1)(x - 6) = 0$
$\therefore (x – 1)(x – 6) = 0$
$\therefore x – 1 = 0 or x – 6 = 0$
$\therefore x = 1$ or$ x = 6$
For $x = 1,$
$f''(1) = 12(1) – 42 = 12 – 42 = – 30 < 0$
$\therefore f ( x )$ attains maximum value at $x =1$.
$\therefore$ Maximum value $= f (1)$
$= 2(1)^3 – 21(1)^2 + 36(1) – 20$
$= 2 – 21 + 36 – 20$
$= – 19 – 20 + 36$
$= – 39 + 36$
$= – 3$
$\therefore$ The function $f ( x )$ has maximum value -3 at $x =1$.
For $x = 6,$
$f''(6) = 12(6) – 42 = 72 – 42 = 30 > 0$
$\therefore f(x)$ attains minimum value at $x = 6.$
$\therefore $ Minimum value $= f(6)$
$= 2(6)^3 – 21(6)^2 + 36(6) – 20$
$= 432 – 756 + 216 – 20$
$= – 128$
$\therefore $ The function f(x) has minimum value – 128 at $x = 6.$

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