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.
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:
Let's look at how we can use these concepts to investigate/describe a shape in more detail.
For the function $f(x)=x^23x4$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:
$f'(x)=2x3$f′(x)=2x−3
Do: The function is increasing where $f'(x)>0$f′(x)>0 :
$2x3$2x−3  $>$>  $0$0 
$x$x  $>$>  $\frac{3}{2}$32 
The function is decreasing where $f'(x)<0$f′(x)<0 :
$2x3$2x−3  $<$<  $0$0 
$x$x  $<$<  $\frac{3}{2}$32 
The function is stationary where $f'(x)=0$f′(x)=0 :
$2x3$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 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.
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 
\ 
. 
/ 
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 
/ 
. 
\ 
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.
Determine the coordinates and nature of any stationary points on the function $f(x)=2x^3+6x^23$f(x)=2x3+6x2−3.
Think: For the function:
$f(x)$f(x)  $=$=  $2x^3+6x^23$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:
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:
Consider the function $f\left(x\right)=\left(x+4\right)\left(x2\right)$f(x)=(x+4)(x−2) drawn below.
What is the $x$xvalue of the stationary point of $f\left(x\right)$f(x)?
What is the region of the domain where $f\left(x\right)$f(x) is increasing?
Write the answer in interval notation.
What is the region of the domain where $f\left(x\right)$f(x) is decreasing?
Write the answer in interval notation.
Consider the function $f\left(x\right)=x^2+4x+9$f(x)=x2+4x+9.
Determine an equation for the gradient function $f'\left(x\right)$f′(x).
Determine the interval in which the function is increasing.
Give your answer in the form of an inequality.
Determine the interval in which the function is decreasing.
Give your answer in the form of an inequality.
Solve for the $x$xcoordinate(s) of the stationary point(s).
If there is more than one, write all of them on the same line, separated by commas.
State the coordinates of the stationary point.
Give your answer in the form $\left(a,b\right)$(a,b).
Consider the four functions sketched below.
Which of the sketches match the information for $f\left(x\right)$f(x) shown in the table? Select all the correct options.
Information about $f\left(x\right)$f(x): 
$f\left(0\right)=5$f(0)=5 
$f\left(2\right)=0$f(−2)=0 
$f'\left(3\right)=0$f′(3)=0 
$f'\left(x\right)>0$f′(x)>0 for $x<3$x<3 
Consider the parabola $y=3x^218x+24$y=3x2−18x+24.
Find $\frac{dy}{dx}$dydx.
Find the $x$xcoordinate of any stationary points.
Complete the table for the value of $\frac{dy}{dx}$dydx around the stationary point to determine its nature.
$x$x  $2$2  $3$3  $4$4 
$\frac{dy}{dx}$dydx  $\editable{}$  $0$0  $\editable{}$ 
Is the point a maximum or a minimum turning point?
Maximum
Minimum
Maximum
Minimum
Find the $x$xintercepts.
Find the $y$yintercept.
Plot the parabola $y=3x^218x+24$y=3x2−18x+24.
Consider the gradient function $f'\left(x\right)=12\left(x+4\right)^2\left(x+7\right)$f′(x)=12(x+4)2(x+7).
Graph the gradient function.
What kind of feature is at the point $\left(7,1617\right)$(−7,−1617) on the graph of $f\left(x\right)$f(x)?
Minimum turning point
Point of inflection
$x$xintercept
Maximum turning point
Minimum turning point
Point of inflection
$x$xintercept
Maximum turning point
What kind of feature is at the point $\left(4,1536\right)$(−4,−1536) on the graph of $f\left(x\right)$f(x)?
Maximum turning point
Minimum turning point
$x$xintercept
Point of inflection
Maximum turning point
Minimum turning point
$x$xintercept
Point of inflection
Use the notations f'(x), f''(x), dy/dx, d^2y/dx^2 [=d/dx(dy/dx)].
Apply differentiation to stationary points.
Use the first and second derivative tests to discriminate between maxima and minima.