Browse Questions

# The smallest value of the polynomial $x^3-18x^2+96x$ in [0,9] is

$(A)\;126\quad(B)\;0\quad(C)\;135\quad(D)\;160.$

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$y=x^3-18x^2+96x$
On differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=3x^2-36x+96$
when $\large\frac{dy}{dx}=0$
$3x^2-36x+96=0$
$\Rightarrow 3(x^2-12x+32)=0$
$\Rightarrow 3(x-8)(x-4)=0$
$\Rightarrow x=8, 4$
Let us consider the points when $x=0, 4, 8, 9$
Step 2
when $x=0$
$y=0$
Step 3
when $x=4$
$y=4^3-18(4)^2+96(4)$
$= 260$
Step 4
when $x=8$
$y=8^3-18(8)^2+96(8)$
$= 120$
Step 5
when $x = 9$
$y=9^3-18(9)^2+96(9)$
$= 135$
Hence the smallest value of the polynomial is 0
Hence B is the correct option.