26.05 Stationary points (Extended)

Connections between a function and its derivative

We have seen we can find the gradient of a function by using its derivative. Can we glean any more information from the derivative function? The applet below shows the graphs of several different functions (linear, quadratic, cubic and the hyperbola $y=\frac{1}{x}$y=1x​ ) together with the graph of their derivative. Explore each function and look for connections between the features of the function and the graph of the derivative such as:

• The steepness of the curve
• Where the function is increasing or decreasing
• Key points such as maximums or minimums
• The shape of the derivative function

Created with Geogebra

What connections did you find? Let's look further at the connections for polynomials and their gradient functions. The shape of a polynomial determines the shape of its derivative, since when we differentiate a polynomial of degree $n$n we obtain a derivative of degree $n-1$n−1. Hence, a linear function will produce a constant derivative, a quadratic function will produce a linear derivative, a cubic function will produce a quadratic derivative and so forth.

Another observation which becomes apparent after looking at a few functions together with their derivatives is that turning points(maximum and minimum) in the function correspond to $x$x-intercepts in the gradient function. Check the graphs below which display this connection. This occurs since the tangent to a turning point will be horizontal and thus have a gradient of zero. (for further discussion see stationary points)

Quadratic	Cubic	Quartic

 

Increasing and decreasing

When a function is increasing, that is as the $x$x values increase the $y$y values increase, this relates to a positive gradient. Thus, when the function is increasing the derivative will be above the $x$x-axis.

When a function is decreasing, that is as the $x$x values increase the $y$y values decrease, this relates to a negative gradient. Thus, when the function is decreasing the derivative will be below the $x$x-axis.

For the graphs below observe where the function is increasing (indicated in green sections), where it is decreasing decreasing (indicated in blue) and where the derivative function is located.

 

Stationary points

A stationary point is where the derivative is zero, that is a point $\left(a,f(a)\right)$(a,f(a)) is said to be a stationary point if $f'(a)=0$f′(a)=0. At this point the tangent is horizontal and thus, the instantaneous rate of change is zero - so the function is momentarily stationary. We have three types of stationary points:

• Local minimum: at this point $f(x)$f(x) has the lowest value in the region of this point. The graph changes from decreasing to increasing at this point. Hence, the derivative will change from negative to positive.
• Local maximum: at this point $f(x)$f(x) has the highest value in the region of this point. The graph changes from increasing to decreasing at this point. Hence, the derivative will change from positive to negative.

• Stationary point of inflection: at this point the gradient does not change either side of the point, it can be either positive either side of the stationary point or negative either side. At points of inflection the rate of change switches from increasing to decreasing or vice versa. So unlike the maximum or minimum it is not the graph itself changing between increasing and decreasing but the gradient of the graph. So the gradient changes from becoming less steep to becoming steeper or vice versa. This will be explored more in further studies of calculus.

Local maximum and minimum are also referred to as turning points, since the function changes between increasing and decreasing at these points.

The reason for the emphasis on "local" minimum or maximum is that they may give the minimum or maximum value within a region of the point but not over the entire domain of the graph. If a point gives the maximum or minimum value over the entire domain of a graph we refer to the point as a global maximum or global minimum. In the first graph below we have both a global and local minimum, as well as a local maximum, however, as the graph is unbounded there is no global maximum. The second graph below shows a restricted domain where the global maximum is in fact at an end-point and not a turning point.

For many functions a local maximum or minimum is also a global maximum or minimum, such as turning points for an unrestricted quadratic graph.

 

Practice questions

Question 1

Question 2

 

 

To find stationary points for a function $y=f(x)$y=f(x):

1. Find the derivative $f'(x)$f′(x)
2. Solve $f'(x)=0$f′(x)=0 for $x$x.
3. Substitute the $x$x-coordinate of any stationary points found back into the original function $f(x)$f(x) to find the $y$y-coordinate of the point.

To determine the nature of a stationary point, we can evaluate the derivative just either side of the stationary point. Thus, if the derivative changes from negative to positive - we have identified a local minimum. If the derivative changes from positive to negative - we have identified a local maximum. And if the sign of the derivative does not change either side of the stationary point - we have identified a stationary point of inflection. A table or diagram indicating the sign and/or shape (increasing, decreasing or constant) is helpful to keep track and visualise the nature of turning points for graphing purposes.

Worked example 

Example 1

For the function $f(x)=\left(x-5\right)\left(x-2\right)\left(x+3\right)$f(x)=(x−5)(x−2)(x+3), find the location and nature of any turning points.

 

Do: Find the derivative and solve $f'(x)=0$f′(x)=0.

Expanding $f(x)$f(x) we obtain:

$f(x)$f(x)	$=$=	$\left(x-5\right)\left(x-2\right)\left(x+3\right)$(x−5)(x−2)(x+3)
	$=$=	$\left(x-5\right)\left(x^2+x-6\right)$(x−5)(x2+x−6)
	$=$=	$x^3-4x^2-11x+30$x3−4x2−11x+30
Hence, $f'(x)$f′(x)	$=$=	$3x^2-8x-11$3x2−8x−11

Solving $f'(x)=0$f′(x)=0:

$3x^2-8x-11$3x2−8x−11	$=$=	$0$0
$\left(3x-11\right)\left(x+1\right)$(3x−11)(x+1)	$=$=	$0$0
$\therefore x$∴x	$=$=	$\frac{11}{3}$113​ or $-1$−1

Substituting these values into $f(x)$f(x) we obtain $f\left(\frac{11}{3}\right)=-\frac{400}{27}\approx-14.8$f(113​)=−40027​≈−14.8 and $f\left(-1\right)=36$f(−1)=36. So we have two stationary points at $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​) and $\left(-1,36\right)$(−1,36). We can use information about the shape of the graph (positive cubic) to ascertain that the point to the left, $\left(-1,36\right)$(−1,36), will be the maximum and the second point, $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​), will be the minimum.

Alternatively, we can use calculus to test the behaviour of the derivative either side and between these two points. Choose convenient $x$x-values before $x=-1$x=−1, (such as $x=-2$x=−2), between $x=-1$x=−1 and $\frac{11}{3}$113​ (such as $x=0$x=0) and after $x=\frac{11}{3}$x=113​ (such as $x=4$x=4). Substitute the values into the gradient function - we are concerned with the sign and whether the function is increasing or decreasing, so you can choose to simply record the sign in the table and not include the value.

$x$x	$-2$−2	$-1$−1	$0$0	$\frac{11}{3}$113​	$4$4
$f'(x)$f′(x)	$17$17	$0$0	$-11$−11	$0$0	$5$5
sign	$+$+	$0$0	$-$−	$0$0	$+$+
shape

The table shows that at $x=-1$x=−1 the graph changes from increasing to decreasing and hence, $\left(-1,36\right)$(−1,36) is a maximum. And at $x=\frac{11}{3}$x=113​ the graph changes from decreasing to increasing and hence, $\left(\frac{11}{3},-\frac{400}{27}\right)$(113​,−40027​) is a minimum.

 

Practice questions

Question 3

Question 4