Find local maxima and local minima for the function \[f(x) = {x^3} - 3x\] .
A.Local max. at \[x = 1\] , local min. at \[x = - 1\]
B.Local max. at \[x = - 1\] local min. at \[x = 1\]
C.Local max. at \[x = - 1\] , no local min.
D.Local min. at \[x = 1\] , no local max.

Hint: Here we have to find the local maxima and local minima for the given function. For that, we will first find the derivative of the function. We will find all the possible points by putting\[f'(x) = 0\]. Then we will find the point of local maxima and local minima. For that, we will further differentiate \[f'(x)\]and then we will put all the points in\[f''(x)\]one by one. If\[f''(x) < 0\], then the point is called as the point of local maxima and if \[f''(x) > 0\], then the point is called as the point of local minima.

Complete step-by-step answer:
Let’s first consider the given function.
\[f(x) = {x^3} - 3x\]
We will find all the points by differentiating the function \[f(x)\]and equating it with zero.
Differentiating function\[f(x)\]with respect to x, we get
\[f'(x) = 3{x^2} - 3\]
Now, we will equate \[f'(x)\] with zero.
\[f'(x) = 0\]
Putting value of \[f'(x)\] in the above equation, we get
\[3{x^2} - 3 = 0\]
We will simplify the equation further.
\[3({x^2} - 1) = 0\]
Now, we will factorize \[({x^2} - 1)\].
\[3(x - 1)(x + 1) = 0\]
We will now apply zero product rule to find all the possible points. So the possible points are :
\[\begin{array}{l}x - 1 = 0\\ \Rightarrow x = 1\end{array}\]
and
\[\begin{array}{l}x + 1 = 0\\ \Rightarrow x = - 1\end{array}\]
So the points are 1 and \[ - 1\].
Now, we will check which point is the point of local maxima and point of local minima. For that, we will differentiate \[f'(x)\] with respect to x.
\[\dfrac{{df'(x)}}{{dx}} = \dfrac{{d(3{x^2} - 3)}}{{dx}}\]
After differentiating both sides with respect to x, we get
\[f''(x) = 6x\]
Now, we will put all the points one by one in equation \[(1)\] .
First putting \[x = 1\] in\[f''(x)\], we get
\[f''(x) = 6 \times 1\]
\[f''(x) = 6\] , which is more than zero i.e. \[f''(x) > 0\]. So this point is the point of local minima.
Therefore,
\[x = 1\] is a point of local minima.
Now, putting \[x = - 1\]in \[f''(x)\]
\[f''(x) = 6 \times - 1\], which is less than zero i.e. \[f''(x) < 0\]. So this point is the point of local maxima.
Thus,
\[x = - 1\]is the point of local maxima.
Therefore, the local minima of function \[f(x)\] is 1 and the local maxima of function \[f(x)\]is -1.
Thus, the correct option is B.

Note: A point is said to be a local maximum point of a function \[f(x)\]at \[x = {x_0}\] when the value of \[f(x)\]at \[x = {x_0}\] is greater than the value of \[f(x)\]at all the points near to the point \[x = {x_0}\]. Similarly, a point is said to be local minimum point of a function \[f(x)\] at \[x = {x_0}\] when the value of \[f(x)\] at \[x = {x_0}\]is smaller than the value of \[f(x)\] at all the points near to the point \[x = {x_0}\].
The common mistake that occurs while solving this question is that sometimes we take the point as the local maximum point when we find \[f''(x) > 0\] after putting that point in \[f''(x)\] , which is wrong. If we get\[f''(x) > 0\] for point \[x = {x_0}\], then that point is a local minimum point but not a local maximum point.