MCQ
The minimum value of the polynomial $x(x + 1) (x + 2) (x + 3)$ is
- A$0$
- B$9/16$
- ✓$- 1$
- D$- 3/2$
$y = (x^2 + 3x) (x^2 + 3x + 2) = z (z + 2)$
$= (z + 1)^2 - 1 = (x^2 + 3x + 1)^2 - 1$
$y_{least} = - 1$ ; this occurs where $z = - 1$ i.e.$ x^2 + 3x + 1 = 0$
or $\frac{{d\,y}}{{d\,x}} = 2 (2x + 3) (x^2 + 3x + 1) = 0 $
$==> x =\frac{{ - \,3\,\, + \,\,\sqrt 5 }}{2} , - \frac{3}{2} or \frac{{ - \,3\,\, - \,\,\sqrt 5 }}{2}$
Here $ x = \frac{{ - \,3\,\, + \,\,\sqrt 5 }}{2}$ & $x = \frac{{ - \,3\,\, - \,\,\sqrt 5 }}{2}$ are the points of local minima and $x = - \frac{3}{2}$ is the point of local maxima . Local maximum value $ =\frac{9}{{16}} $ 
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