Question

Find the minimum and maximum value of the function $y=x^3-3x^2+6$. Find the values of x at which it occurs.

• A. Maxima at x = 0, Maximum value = 6
• B. Maxima at x = 2, Maximum value = 2
• C. Minima at x=0,minimum value = 6
• D. Minima at x=2,minimum value = 2

Answer 1

Answer: (A), (D)

The correct options are
A

Maxima at x = 0, Maximum value = 6

D

Minima at x=2,minimum value = 2

Given $y=x^3-3x^2+6$
Differentiating y w.r.t. 'x', $\frac{dy}{dx}=3x^2-6x$
Putting $\frac{dy}{dx}=0$, we will get the values at which function is maximum or minimum
$\Rightarrow 3x^2-6x=0\Rightarrow x(3x-6)=0$
$\Rightarrow x=0,+2$
To distinguish values of x as the point of maximum or minimum, we need 2nd derivative of the function.
$\therefore \frac{d^{2}y}{d{x}^2}=6x-6;Now(\frac{d^{2}y}{d{x}^2}{)}_{x=0}=-6<0$
$\therefore Atx=0\Rightarrow \text{It is maximum}{\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=2}=6\left(2\right)-6=6>0\therefore Atx=2\Rightarrow \text{It is minimum}$
$Hencex=0isapointofmaximumandx=2isapointofminimumSo,maximumvalueofy=0^3-3.0+6=6minimumvalueofy=(2{)}^3-3(2{)}^2+6=2$