MCQ
The minimum value of the function $y = 2{x^3} - 21{x^2} + 36x - 20$ is
- ✓$-128$
- B$-126$
- C$-120$
- DNone of these
$f'(x) = 6{x^2} - 42x + 36$
Put $f'(x) = 0$==> $6{x^2} - 42x + 36 = 0$==> ${x^2} - 7x + 6 = 0$
==> ${x^2} - 6x - x + 6 = 0$==> $(x - 1)(x - 6) = 0$==> $x = 1,\,6$
Now, $f''\,(x) = 12x - 42$
$f''\,(1) = - 30 = - ve$ and $f''\,(6) = 30 = + ve$
Hence $x = 6$ is the point of minima
$\therefore$ Minimum value = $f(6) = 2{(6)^3} - 21{(6)^2} + 36 \times 6 - 20$
$\therefore$ $f(6) = - 128$.
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