Stationary points are particularly comes under calculus which is the main branch of mathematics. It is also called critical point. Stationary points of a differential functions is nothing but one variable point of the domain of that function where the slope of the graph is zero. It is a input values of differential functions where all partial derivatives are zero. Stationary points also called turning point. It can be visualize by given graph of the function. At this point the function tends to change its direction. |

Here in the above graph we can easy notice points A,B,C referred stationary points. A, B, C points are drawn tangent on the graph parallel to the X axis or horizontal if the function is with one variable . In the case of two variables, points A, B, C are correspond to the tangent plane drawn parallel to the XY plane or horizontal.

The word stationary point is little different from the critical points or turning point.

Critical point is tend to infinity or undefined, where as stationary point is tend to zero. On the other hand turning point tend to change the sign. There points may referred as stationary points in case of differentiation.(But not all the turning points are stationary points).

The differentiation of stationary points can be done in many way and stationary points itself are many types. To get gradient as zero be need to know double differentiation of function. The method of twice differentiation is as follows;

$\frac{d}{dx}$ $(\frac{df}{dx})$ = $\frac{d^{2}f}{dx^{2}}$ = $f"(x)\ =\ f^{(2)}(x)$

The nature of stationary points can be determine from their minimum points, maximum points and inflection of points. To find the nature of points second derivative is one of the easiest method. The curve of stationary points is exist when $\frac{dy}{dx}$ = $0$ Stationary points nature is described as follows:

**1)**The maximum turning point at point x is exists, When first derivative $f '(x)\ =\ 0$ and second derivative $f ''(x)\ <\ 0$.

**2)**The minimum turning point at point x is exists, when first derivative $f '(x)\ =\ 0$ and second derivative $f ''(x)\ >\ 0$.

**3)**The point of inflection is exists, when $f '(x)\ =\ 0$ and $f ''(x)\ =\ 0$.

The stationary points of a function describes the working method of function. It is important to find the function. This is useful in sketching the function, to track on function where it changes direction.

There are mainly three types of stationary points given as follows,

**1)**Maximum point.

**Minimum point.**

2)

2)

**Inflection point.**

3)

3)

Below find some steps to find the stationary points of functions.

**1)**Compute the given function $f(x)$ by first derivative $f'(x)$.

**2)**Find the value of $x$ by substituting $f'(x)$ value as zero.

**3)**Find the value of $y\ =\ f(x)$ by substituting $x$ value from step $2$.

**4)**$(x,\ y)$ given the stationary points.

**5)**Find the value of $f"(x)$ at each points to determine the stationary points nature.

**Example 1:**

Calculate the stationary points for the function $f(x) = x^{3} - 6 x^{2} + 9 x + 1$ and the nature of each stationary point.

**Solution:**

$f(x) = x^{3} - 6 x^{2} + 9 x + 1$

$f '(x) = 3 x^{2} - 12 x + 9$

For stationary points

$f '(x) = 0$

$3 x^{2} - 12 x + 9 = 0$

$x^{2} - 4 x + 3 = 0$

$(x - 1) (x - 3) = 0$

$x = 1 , 3$

$y = f(x) = x^{3} - 6 x^{2} + 9 x + 1$

$f(1) = (1)^{3} - 6 (1)^{2} + 9 (1) + 1$

= $5$

$f(3) = (3)^{3} - 6 (3)^{2} + 9 (3) + 1$

= $1$

Therefore, the stationary points are $(1, 5)$ and $(3, 1)$.

Now, $f "(x) = 6 x - 12$

$f "(1) = 6 - 12 = -6$ (negative)

Thus, there is a maximum at $(1, 5)$.

$f "(3) = 18 - 12 = 6$ (positive)

Thus, there is a minimum at $(3, 1)$.

**Example 2:**

Determine the stationary points and their nature for the curve $y = f(x) = 2 x^{3} - 4$.

**Solution :**

For stationary point, we have

$f '(x) = 0$

$6 x^{2} = 0$

$x = 0$

$y = f(0) = -4$

Thus, $(0, - 4)$ is a stationary point.

In order to find the nature of stationary point, let us find $f”(x)$

$f ”(x) = 12 x$

at $x = 0, f ”(x) = 0$

We know that if $f ”(x) = 0$, then there is a possible point of inflection. So, a point of inflection lies at $(0, -4)$.

Now, we will check for the concavity at this point

for $x = - 1$ which is less than $0$, we have

$f”(x) = 12(-1) < 0$

for $x = 0$, we have

$f"(x) = 0$

for $x = 1$ which is greater than $0$, we have

$f”(x) = 12(1) > 0$

Here, the the concavity of the curve is changing at the point $(0, -4)$, hence there is a horizontal point of inflection.