Reviewing the lesson before this, we learned what the derivative indicates regarding the direction a curve is going at a given point. The derivative tells us what the slope of the tangent to a certain point is. We can use that value to shape our curves.

We do this by first identifying the possible Local Extremes and then looking at the values of the derivative “around them”. This may seem abstract but will make sense once we work through some visual examples.

[Reminder: a Local Extreme is the same as a Maximum or Minimum]

Recall that, for a function of *x*, f(x), if we have the point a in that function where,

f’(a) = 0 or f’(a) = undefined,

Then what we do is look at the values of the derivative “around it” to tell us if our point at a, is a Local Extreme or not, and if it is a Local Extreme, whether its a Local Maximum or Local Minimum.

In terms of derivatives and local extremes,

*We have a Minimum*, if for the point *p*, where f’(p) = 0, and the points around it, i.e. the point to the left p - 1 and the point to the right p + 1,

If, f’(p - 1) = **negative** and f’(p + 1) = **positive**

*We have a Maximum*, if for the point *p*, where f’(p) = 0, and the points around it, i.e. the point to the left p - 1 and the point to the right p + 1,

If, f’(p - 1) = **positive** and f’(p + 1) = **negative**

*We do not have a Maximum or Minimum*, if for the point *p*, where f’(p) = 0, and the points around it, i.e. the point to the left p - 1 and the point to the right p + 1,

If, f’(p - 1) = **positive** and f’(p + 1) = **positive**

Or

If, f’(p - 1) = **negative** and f’(p + 1) = **negative**

Once we see examples, it will make more sense!

Below we have the graph of a function of *x*, f(x).

f’(a) = 0

f’(b) = 0

f’(c) = 0

If we remember, we can determine if the points *a*, *b*, and *c* are local extremes (maximums or minimums) by looking at the shape of the curve “around it”.

To review this non exact method of “looking” at the graph to determine maximums and minimums, let's walk through the steps and then introduce our mathematical language,

At points a, b, and c, we have points where the gradient is equal to zero, i.e. where there is a possible local extreme, in mathematical language, we can write it as,

At a, if we look at each side of the point, to the left at a - 1 and to the right at a +1, we will see the value of the slope,

At a - 1 the slope is a negative slope, this means that the derivative of the function at the point x = a - 1 will be **negative**

Mathematically, we write this **negative** slope at a - 1 as,

f’(a - 1) = **negative**

At a +1, the slope is a positive slope, this meant that the derivative of the function at the point x = a +1 will be **positive**

Mathematically, we write this **positive** slope at a +1 as,

f’(a +1) = **positive**

Conclusion: as we have the point a being a possible local extreme, with f’(a) = 0, and all the points to the left and right of a (a - 1 & a +1) being negative and positive slopes respectively, we can confirm a is the lowest local point and a is a minimum.

Looking at this on a graph we see the points a - 1 & a +1 and their negative and positive slopes.

Now let's repeat this process for point b,

We look at the points that are “around” b, i.e. the points just to the left and to the right of b, this will be b - 1 and b + 1,

We can see that to the left of b, b - 1, that the slope is positive,

f’(b - 1) = **positive**

We can see that to the right of b, b + 1, that the slope is also positive,

f’(b + 1) = **positive**

So, although b is a possible local extreme with f’(b + 1) = 0, it is not a local extreme (maximum or minimum) as it is not either the lowest or highest value amongst the points “around it”.

Look below at the illustration to see how the point b is not a maximum or minimum, demonstrated by the positive gradient to the left and right of the point,

Lastly, for point c,

We look at the points that are “around”* c*, i.e. the points just to the left and to the right of *c*, this will be c - 1 and c + 1,

We can see that to the left of c, c - 1, that the slope is positive,

f’(c - 1) = **positive**

We can see that to the right of *c*, c + 1, that the slope is a negative,

f’(c + 1) = **negative**

From these findings on the gradients of the slopes to the right and to the left of the point c, we can conclude that c, is a maximum, as its the highest value compared to the points around it.

We can see in the visual, that by showing that if the gradient to the left of a possible local extreme is positive and if the gradient to the right is negative, then that possible local extreme is a Maximum,

Question: For the possible local extremes a, b, and c marked on the graphed function f(x) below,

i) Mark where the points “around” that point would be i.e. to the left (point - 1) and to the right (point + 1)

ii) Then identify if the gradient is positive or negative

iii) Therefore determine whether that point (*a*, *b*, and *c*) is a maximum, minimum or neither.

[Remember, to the left or to the right does not have to be exactly one value to the left or two the right, this method not exact and is allowed some leniency]

So we have our points, a, b, and c and we need to mark the points to the left and right of each of these points. We roughly do this as shown below.

Now we have our points "around" a, b, and c marked

For part ii), let's look at the graph and determine what the slope is.

a - 1 - the line is moving up therefore the gradient is positive. i.e. f '(a - 1) =

a + 1 - the line is also moving up therefore the gradient is also positive. i.e. f '(a + 1) =

b - 1 - the line is moving up therefore the gradient is positive. i.e. f '(b - 1) =

b + 1 - the line is moving down therefore the gradient is negative. i.e. f '(b + 1) =

c - 1 - the line is moving down therefore the gradient is negative. i.e. f '(c - 1) =

c + 1 - the line is moving up therefore the gradient is positive. i.e. f '(c + 1) =

Now part iii) - Determine if our points, a, b, and c are maximums, minimums, or neither.

At a, a - 1 and a + 1 are both positive values for the derivative. i.e. they are both moving up. This means that the point point *a*, is not the largest or smallest value in that area and is therefore **not a maximum or a minimum**.

At b, b - 1 and b + 1 are positive and negative gradients respectively. This means that *b* is the highest value of f(x) in that area and therefore it **is a maximum**.

At c, c - 1 and c + 1 are negative and positive respectively. This means that *c* is the lowest value of f(x) in that area and is therefore **a minimum**.

We have demonstrated how we can identify areas of the curve by finding the values where the derivative of the function is equal to zero (or in some cases where the derivative is undefined). When we find those points, we can then look at the value of the derivative “around” these points, when we do this, it tells us the values of the slopes of the curve “around” our flat points. We can use these derivatives to start to roughly shape curves of complex functions.

Below, we will learn how to roughly draw the curve of a function simply by using the derivative of that function.

This is the first step in being able to illustrate the graph of a function. We should be familiar with it, what it involves is, identifying the parts of the function f(x), where f’(x) = 0 or f’(x) is undefined.

We do this by calculating the derivative of our given function, then equate that function to zero, and solve for the values of x that make that function equal to zero.

This is identifying the parts of the function that are “flat” i.e. where f’(x) = 0. These values are the possible local extremes.

Once, we have our x-value/s for our flat point/s, input that x-value into our original function, f(x), to give us the value of the y-value.

This is similar to the example we did earlier in this lesson. We look at the points that are “around” our flat points. This is not an exact method, we need to roughly determine what values are “close” to the flat points.

So, say we have found the the flat point for a function is x = 2 , where f’(2) = 0, previously, we have just been using symbols like a - 1 and a + 1, but when we start to do real examples and have flat points like x = 2, it is to our discretion how we pick the points “around” x = 2, in most cases we are fine to use +1 and -1. For this example, we could pick the points “around” x = 2 as 1 and 3, but for argument sake, we could also pick x = 2.1 and x = 1.9

When we have more than one flat point, we do not need to pick two points between two or more flat points (a tad confusing but the example may help). For example, if we had two flat points of,

x = 3 and x = 6,

for the points to the right of x = 3 and to the left of x = 6, we only need to pick **one value between 3 and 6**.

So, instead of picking, say, x = 4 for our value to the right of 3 and x = 5 for our value to the left of x = 6, we only need to pick one value,

So we could choose any value between x = 3 and x = 6. Remember the purpose of the points “around” our flat points is to give us an idea of what the curve or the gradient is like between the flat points.

Once we have chosen our left and right x-values for the points “around” our flat points, input these x-values into our original function to give us their y-value as well.

Now we have determined where our flat points are, and what the points “around” them are, we need to determine the gradient of the points “around” the flat points. For the purposes of roughly drawing our graph, we only need to know if the gradient is positive or negative.

A reminder, we determine the gradient of a point on a function f(x), but inputting that point into the derivative of that function.

So, as per the example we used in the previous Step, if we had a flat point x = 2 then to determine the gradients of the points around it, we would input those values into the derivative, then determine if the gradient is positive or negative. If it is positive, we know that on that side, that the graph is going up, and if it is negative the graph is going down.

So say for example, we inputted x = 1 into f’(x),

f’(1) = 3 → this means we have a positive gradient and therefore the graph will be going up at that point

Then for the point to the right of x = 2, being x = 3,

f’(3) = -2 → this means we have a negative gradient and therefore the graph will be going down at that point.

After we have determined our flat points, determined our points “around” them, determined what the gradients of those close points are, the next step is to lie out those flat points and points around them and then sketch a flat, upwards, or downwards line underneath those points. Don't worry if this seems strange, it will make more sense when we see examples. It is the new concept we introduce in this lesson, and it is where our mathematics start to come together into something visual.

So, once we have completed step three (determining what the gradients around our flat points are), we will now sketch either an upwards line / for our positive gradients, or a downwards line \ for our negative gradients. Also, for our flat points, we sketch at flat line _ to represent the flat tangent line at that point. Again, this will make more sense once we see the visual below.

In this lesson, we are working backwards from the way we are used to. We will be given an equation and then calculate its flat points and gradients around the flat points, and then use our upwards / , downwards \ and flat _ lines to outline what our graph will look like.

This last step simply involves plotting our roughly drawn straight line gradients on a graph, in line with our flat point/points and the points “around” them.

First, plot each point on a x and y graph. Then,

Sketch either the upwards / or downwards \ or flat _ line each point, as if you are sketching the tangent of a point on a curve.

Lastly, connect the ends of each of those plotted straight lines, to come up with a rough representation of what the curve will look like.