Question

Find the local maxima and minima of the function $y=x^3-3x+2$.

• A. $6$ , $3$
• B. $4$ , $0$
• C. $8$ , $4$
• D. $0$ , $4$

Answer 1

The correct option is B $4$ , $0$

Given function $y=x^3-3x+2$
$\frac{dy}{dx}=3x^2-\mathrm{3.......}\left(1\right)$
At stationary points, $\frac{dy}{dx}=0$,
$\Rightarrow 3x^2-3=0$
$\Rightarrow 3(x^2-1)=0$
$\Rightarrow (x-1)(x+1)=0$
$\Rightarrow x=1,x=-1$

Differentiating the equation on both sides again,
$\frac{d^{2}y}{d{x}^2}=6x.....\left(2\right)$
Putting $x=1$ in equation $\left(2\right)$
$\frac{d^{2}y}{d{x}^2}=6>0$
$\Rightarrow$ At point $x=1$ , local minimum occurs
and local minimum is given by
$y\left(1\right)=(1{)}^3-3(1)+2=0$
Now, putting $x=-1$ in equation $\left(2\right)$
$\frac{d^{2}y}{d{x}^2}=-6<0$
So, at point $x=-1$, local maximum occurs and local maximum is given by $y(-1)=(-1{)}^3-3(-1)+2=4$
So, local maximum and minimum are $4$ and $0$, respectively