# STATIONARY POINTS (MINIMUM, MAXIMUM AND INFLECTION POINTS)

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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