11.08 Using the first derivative

Having practised using all of our available rules for differentiation, we will look at how we can use the first derivative to understand the shape of a function. The first derivative reveals key information about important points known as local minimums, maximums and points of inflection. This will enable us to develop a more accurate representation of a function than previously possible.

Increasing, decreasing and stationary points

We have previously encountered the following terminology related to the gradient of functions:

Gradient	$f(x)$f(x) is:	$f'(x)$f′(x)
positive	increasing	$f'(x)>0$f′(x)>0
negative	decreasing	$f'(x)<0$f′(x)<0
$0$0	stationary	$f'(x)=0$f′(x)=0

If the gradient is $0$0 the function is considered to be stationary at that point, that is, it is neither increasing or decreasing. Note that stationary points usually signify a change in either the sign or nature of the gradient and can be very useful in defining the shape of a function.

For example, the function in the diagram below is:

• increasing at $C$C and $G$G
• decreasing at $A$A and $E$E
• stationary at $B$B, $D$D and $F$F.

Let's look at how we can use these concepts to investigate/describe a shape in more detail.

Worked examples

Example 1

For the function $f(x)=x^2-3x-4$f(x)=x2−3x−4, determine the domain for which curve is increasing, decreasing and stationary.

Think: We first need to find the first derivative of the function:

$f'(x)=2x-3$f′(x)=2x−3

Do: The function is increasing where $f'(x)>0$f′(x)>0 :

$2x-3$2x−3	$>$>	$0$0
$x$x	$>$>	$\frac{3}{2}$32​

 

The function is decreasing where $f'(x)<0$f′(x)<0 :

$2x-3$2x−3	$<$<	$0$0
$x$x	$<$<	$\frac{3}{2}$32​

 

The function is stationary where $f'(x)=0$f′(x)=0 :

$2x-3$2x−3	$=$=	$0$0
$x$x	$=$=	$\frac{3}{2}$32​

 

Let's look at a sketch of the function and its associated gradient function:

Stationary points

Stationary points can be further classified into three different types as follows–minimum and maximum turning points, and stationary or horizontal points of inflection. We will now look at each stationary point in more detail.

Minimum turning point

A minimum turning point is where the gradient changes from negative to positive moving left to right or another way to say this is the function changes from decreasing to increasing.

The behaviour of the derivative either side a local minimum can be summarised as follows:

$x$x	LHS	Minimum	RHS
$f'(x)$f′(x)	$f'(x)<0$f′(x)<0	$f'(x)=0$f′(x)=0	$f'(x)>0$f′(x)>0
	\	.	/

Maximum turning point

A maximum turning point is where the gradient changes from positive to negative moving left to right or another way to say this is the function changes from increasing to decreasing.

The behaviour of the first derivative either side a local maximum can be summarised as follows:

$x$x	LHS	Maximum	RHS
$f'(x)$f′(x)	$f'(x)>0$f′(x)>0	$f'(x)=0$f′(x)=0	$f'(x)<0$f′(x)<0
	/	.	\

Stationary or horizontal point of inflection

A stationary or horizontal point of inflection is where the gradient equals $0$0 but the function is either increasing or decreasing on both sides of this point.

The behaviour of the first derivative either side a stationary point of inflection can be summarised as follows:

$x$x	LHS	Stationary point of inflection	RHS
$f'(x)$f′(x)	$f'(x)>0$f′(x)>0	$f'(x)=0$f′(x)=0	$f'(x)>0$f′(x)>0
	/	.	/
$f'(x)$f′(x)	$f'(x)<0$f′(x)<0	$f'(x)=0$f′(x)=0	$f'(x)<0$f′(x)<0
	\	.	\

Let's look at how we can use these ideas to better understand the nature, shape and important points of functions.

Worked examples

Example 2

Determine the coordinates and nature of any stationary points on the function $f(x)=2x^3+6x^2-3$f(x)=2x3+6x2−3.

Think: For the function:

$f(x)$f(x)

$=$=

$2x^3+6x^2-3$2x3+6x2−3

 

We want to find the derivative, and then determine the values of $x$x for which the derivative is zero. Then we can observe the gradient either side of any stationary points to determine their nature.

Do: Taking the first derivative:

$f'(x)$f′(x)

$=$=

$6x^2+12x$6x2+12x

 

To find any stationary points we solve for $f'(x)=0$f′(x)=0:

$f'(x)$f′(x)	$=$=	$6x^2+12x$6x2+12x
$0$0	$=$=	$6x^2+12x$6x2+12x
$6x(x+2)$6x(x+2)	$=$=	$0$0
$x$x	$=$=	$0,-2$0,−2

 

Therefore, there are stationary points at $x=0,-2$x=0,−2. Let's now look at the behaviour of the first derivative to the left and right of these points to determine whether the stationary points are a local minimum, maximum or point of inflection.

For $x=0$x=0:

$x$x	$-1$−1	$0$0	$1$1
$f'(x)$f′(x)	$-6$−6	$0$0	$18$18
	\	.	/

Therefore, as the gradient changes from negative to positive, that is, the function goes from decreasing to increasing, there is a local minimum at $x=0$x=0.

For $x=-2$x=−2:

$x$x	$-3$−3	$0$0	$-1$−1
$f'(x)$f′(x)	$18$18	$0$0	$-6$−6
	/	.	\

Therefore, as the gradient changes from positive to negative, that is, the function goes from increasing to decreasing, there is a local maximum at $x=-2$x=−2.

A sketch of this function and its gradient function confirms what we have found:

 

Using the first derivative

Knowing what we know about how a derivative can help us identify key features of the original function, we can now move on to sketching functions from the derivative information.

To sketch a function we need to know:

• The general shape the function will have. Knowing the degree of the function will tell us the general shape and end behaviour. The first derivative is always one degree less than the original function.
• For polynomial functions, the sign of the coefficient of the first derivative will match the sign of the original function. So if the first derivative is a negative quadratic, then the original function is a negative cubic. This tells us the general shape.
• The zeros of the first derivative will tell us where the original function has a gradient of zero (horizontal tangent).
• If we check for the sign of the gradient at points either side of the stationary points we can piece together the general shape of the curve.
• $x$x- and $y$y-intercepts can be found by using the original function.

Practice questions

Question 1

Question 2

Question 3

Question 4

Question 5