A stationary point is a point on a curve where the gradient is equal to zero.

Symbolically, the stationary point is where

$$\frac{dy}{dx}=0$$

In other words, the derivative function of the curve equals zero at a stationary point.

It is important to note here that stationary points are also known as **critical points **

There are three types of stationary points viz: **Minimum, maximum, and inflection point**.

**The maximum point of a curve or function** is the point where the curve changes from **positive to negative **and the gradient such point is zero.

In other words, the maximum point is the point where the graph changes from **increasing to decreasing (rising to falling).**

The point where the gradient is zero and the curve changes from negative to positive is called the **minimum point of the curve or function**.

In other words, the minimum point is the point where the graph changes from decreasing to increasing (falling to rising)

For example, in the curve below, P Is the maximum point of the curve because the gradient of the curve changes from positive to negative.

Q Is the minimum point of the curve because the gradient of the curve changes from negative to positive.

Because there is always a "turning around" at the maximum and minimum point, Both are collectively called **turning points.**

It is possible to have a turning point, the gradient on either side of which is the same, such a point is called an **inflection point**.

An inflection point is the point of **changes in concavity**. That is, the inflection point is the point where concave changes to convex or convex changes to concave.

**PROCEDURES FOR FINDING AND DETERMINING STATIONARY POINTS**

1. Given a function $y=f(x)$

2. Take the derivative and equate it to zero

$$\frac{dy}{dx}=0$$

3. Substitute the value(s) of x in the original equation $y=f(x)$ to obtain the corresponding coordinate of the stationary points

4. Take the second derivative $\frac{dy}{dx}$ and substitute the value(s) of x obtained in ii).

5. If the result is:

A) **Positive**, Then it's is a **minimum point**

B) **Negative**, It's a **maximum point**

C) **Zero or undefined**, It's an **inflection point**

__Example 1__

Locate the turning point on the curve $y=5x^2+4x$ and determine its nature by examining the sign of the gradient.

**solution:**

$y=5x^2+4x$

First, we take the derivative via power rule

$$\frac{dy}{dx}=10x+4$$

Let's now equate it to zero

$$10x+4=0$$

$$10x=-4$$

$$x=\frac{-4}{10}=\frac{-2}{5}=-0.4$$

To obtain the corresponding coordinate of the statutory point, we insert the value of x in the original expression; $y=5x^2+4x$

$$y=5\left(\frac{-2}{5}\right)^2+4 \left(\frac{-2}{5}\right)$$

$$y=5\left(\frac{4}{25}\right)-\frac{8}{5}$$

$$y=\frac{20}{25}-\frac{8}{5}$$

$$y=\frac{20-40}{25}$$

$$y=\frac{-20}{25}=\frac{-4}{5}=-0.8$$

Hence, the statutory point is (-0.4, -0.8)

To determine the nature of the statutory point, We take the second derivative of the function using the idea of successive differentiation.

if $\frac{dy}{dx}=10x+4$, then

$$\frac{d^2y}{dx^2}=10$$

Because the result (10) is positive, (-0.4, -0.8) is a **minimum point**

__Example 2__

Locate the turning point on the curve $y=12\sqrt{x}+x^{\frac{3}{2}}$ and determine its nature by examining the sign of the gradient.

**Solution**

The above expression can be rewritten as

$y=12x^{\frac{1}{2}}+x^{\frac{3}{2}}$

Now, let take the first derivative

$$\frac{dy}{dx}=6x^{\frac{-1}{2}}-\frac{3x^{\frac{1}{2}}}{2}$$

Next, we equate it to zero

$$6x^{\frac{-1}{2}}-\frac{3x^{\frac{1}{2}}}{2}=0$$

$$6x^{\frac{-1}{2}}=\frac{3x^{\frac{1}{2}}}{2}$$

Cross multiplying

$$12x^{\frac{-1}{2}}=3x^{\frac{1}{2}}$$

$$\frac{12x^{\frac{-1}{2}}}{x^{\frac{-1}{2}}}=\frac{3x^{\frac{1}{2}}}{x^{\frac{-1}{2}}}$$

$$12=3x^{\left(\frac{1}{2}-(\frac{-1}{2})\right)}$$

$$12=3x$$

$$4=x$$

$$x=4$$

To obtain the corresponding coordinate, we insert the value of x in the original function

$$y=12\sqrt{4}+4^{\frac{3}{2}}$$

$$y=12(2)-8$$

$$y=16$$

Hence, the stationary point is (4, 16)

To determine the nature of the stationary, we take the second derivative of the function

$$\frac{d^2y}{dx^2}=\frac{-3}{x^{\frac{3}{2}}}-\frac{3}{4(x)^{\frac{1}{2}}}$$

Now, let's insert the value of x

$$\frac{-3}{(4)^{\frac{3}{2}}}-\frac{3}{4(4)^{\frac{1}{2}}}$$

$$\frac{-3}{8}-\frac{3}{8}$$

$$\frac{-6}{8}=-0.75$$

Since $-0.75$ is negative, (4, 16) is the **maximum point of the curve.**

__Example 3__

Find the minimum and maximum point on the curve $y=x^3+3x^2-24x$ .

**Solution**

$y=x^3+3x^2-24x$

First, we need to obtain the derivative

$$\frac{dy}{dx}=3x^2+6x-24$$

Now, let's equate it to zero

$$3x^2+6x-24=0$$

For simplicity, let's divide through by 3.

$$\frac{3x^2}{3}+\frac{6x}{3}-\frac{24}{3}=\frac{0}{3}$$

$$x^2+2x-8=0$$

$$x^2+4x-2x-8=0$$

$$x(x+4)-2(x+4)=0$$

$$x=2, x=-4$$

Unlike the previous examples, we have two values. This means this curve has both a minimum and maximum point. So, we insert the values of x in the original function

When x=2,

$$y=(2)^3+3(2)^2-24(2)$$

$$y=8+12-48$$

$$y=-28$$

When x=-4,

$$y=(-4)^3+3(-4)^2-24(-4)$$

$$y=-64+48=96$$

$$y=-80$$

The coordinates of the turning points are (2, -28) and (-4, 80).

Now, we need to determine the nature of the stationary point. hence, we take the second derivative

if $\frac{dy}{dx}=3x^2+6x-24$, then

$$\frac{d^2y}{dx^2}=6x+6$$

Now, we insert the value of x in $6x+6$

When x=2

$$6(2)+6=18$$

Because 18 is positive, (2, -28) is the minimum point of the curve

When x=-4

$6(-4)+6=-18$

Because -18 is negative, (-4, 80) is the maximum point of the curve.

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