We have seen we can find the gradient of the tangent to a curve 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:
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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 (maximums and minimums) in the function correspond to $x$x-intercepts in the gradient function. Check the graphs below to confirm this connection. This occurs because the tangent at a turning point will be horizontal and thus have a gradient of zero.
Quadratic | Cubic | Quartic |
When a function is increasing, that is as the $x$x values increase the $y$y values increase, the tangent to the curve has a positive gradient. Thus, when the function is increasing the derivative will be above the $x$x-axis (positive).
When a function is decreasing, that is as the $x$x values increase the $y$y values decrease, the tangent to the curve will have a negative gradient. Thus, when the function is decreasing the derivative will be below the $x$x-axis (negative).
For the graphs below observe where the function is increasing (indicated in green sections), where it is decreasing (indicated in blue) and where the derivative function is located.
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 gradient of the tangent is zero - so the function is momentarily stationary. We have three types of stationary points:
Stationary point of inflection: at this point the gradient does not change either side of the point, it can be positive both sides of the stationary point or negative both sides of the point. At points of inflection the steepness of the gradient of the graph switches from increasing to decreasing or vice versa. 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.
Consider the function $y=x^2-1$y=x2−1 drawn here.
Which of the following graphs represent $y'$y′?
Consider the gradient function $f'\left(x\right)$f′(x) drawn here. Which of the following graphs are possible representations of the original function $f\left(x\right)$f(x)?
Select the two that apply.
Recall that we can sketch the general shape of a given polynomial using facts about:
Previously we had to use technology to find the location and nature of stationary points and where a function is increasing or decreasing. We can now use calculus and the derivative to achieve this.
To find stationary points for a function $y=f(x)$y=f(x):
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.
$n$n | $a>0$a>0 | $a<0$a<0 |
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Even |
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Odd |
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), determine:
(a) The $x$x-intercepts.
(b) The $y$y-intercept.
(c) The location of any turning points.
(d) The nature of the turning points.
(e) Use the information to sketch the function.
Think: This is cubic and if we expanding the brackets we would have a positive leading coefficient. This tells us the general shape with the tails of the function going from bottom left to top right.
(a) Determine the $x$x-intercepts.
Think: This is when $y=0$y=0 so solve $f(x)=0$f(x)=0:
$\left(x-5\right)\left(x-2\right)\left(x+3\right)=0$(x−5)(x−2)(x+3)=0
As the function is in factored form we can see the solutions are $x=5$x=5, $2$2 and $-3$−3. So the $x$x-intercepts are $\left(5,0\right)$(5,0), $\left(2,0\right)$(2,0) and $\left(-3,0\right)$(−3,0).
(b) Determine the $y$y-intercept.
Think: This is when $x=0$x=0:
$f\left(0\right)$f(0) | $=$= | $\left(0-5\right)\left(0-2\right)\left(0+3\right)$(0−5)(0−2)(0+3) |
$=$= | $30$30 |
Thus, the $y$y-intercept is $\left(0,30\right)$(0,30).
(c) Determine the location of the turning points.
Think: Calculate 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).
(d) Determine the nature of the turning points.
Think: There are two methods of doing this. Remember to show working and reasoning with whichever method you decide to use:
Method 1 - shape of the graph:
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.
Method 2 - sign test:
Alternatively, we can check the sign 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$4 |
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$f'(x)$f′(x) | $17$17 | $0$0 | $-11$−11 | $0$0 | $5$5 |
sign |
$+$+ | $0$0 | $-$− | $0$0 | $+$+ |
shape |
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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.
(e) Plotting the points we have so far and sketching a smooth curve connecting them we obtain the graph:
Consider the function $f\left(x\right)=3x^2-54x+241$f(x)=3x2−54x+241
Determine an expression for the derivative of $f\left(x\right)$f(x).
Hence solve for the $x$x-coordinate of the stationary point.
Complete the following table of values.
$x$x | $8$8 | $9$9 | $10$10 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Which of these statements is correct?
$\left(9,f\left(9\right)\right)$(9,f(9)) is a minimum turning point.
$\left(9,f\left(9\right)\right)$(9,f(9)) is a maximum turning point.
Consider the function $f\left(x\right)=\left(x^2-9\right)^2+4$f(x)=(x2−9)2+4.
Expand and simplify the expression for $f\left(x\right)$f(x).
Determine an expression for the derivative of $f\left(x\right)$f(x).
Hence locate the $x$x-values of the stationary points.
Enter your solutions on the same line separated by a comma.
Complete the following table of values.
$x$x | $-4$−4 | $-3$−3 | $-2$−2 |
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$f'\left(x\right)$f′(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
$x$x | $-1$−1 | $0$0 | $1$1 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
$x$x | $2$2 | $3$3 | $4$4 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Which of these statements is correct?
$\left(-3,f\left(-3\right)\right)$(−3,f(−3)) is a minimum turning point.
$\left(-3,f\left(-3\right)\right)$(−3,f(−3)) is a maximum turning point.
Select the correct statement.
$\left(0,f\left(0\right)\right)$(0,f(0)) is a minimum turning point.
$\left(0,f\left(0\right)\right)$(0,f(0)) is a maximum turning point.
Consider the function $f\left(x\right)=\left(x^2-4\right)^2+4$f(x)=(x2−4)2+4.
State the coordinates of the $y$y-intercept.
Give your answer in the form $\left(a,b\right)$(a,b).
Determine an equation for $f'\left(x\right)$f′(x).
Hence solve for the $x$x-coordinate(s) of the stationary point(s).
If there is more than one, write all of them on the same line separated by commas.
By completing the tables of values, find the gradient of the curve at $x=-2.5$x=−2.5, $x=-1.5$x=−1.5, $x=-0.5$x=−0.5, $x=0.5$x=0.5, $x=1.5$x=1.5 and $x=2.5$x=2.5.
$x$x | $-2.5$−2.5 | $-2$−2 | $-1.5$−1.5 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
$x$x | $-0.5$−0.5 | $0$0 | $0.5$0.5 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
$x$x | $1.5$1.5 | $2$2 | $2.5$2.5 |
---|---|---|---|
$f'\left(x\right)$f′(x) | $\editable{}$ | $0$0 | $\editable{}$ |
Select the correct statements.
$\left(2,4\right)$(2,4) is a maximum turning point.
$\left(0,20\right)$(0,20) is a maximum turning point.
$\left(-2,4\right)$(−2,4) is a minimum turning point.
$\left(0,20\right)$(0,20) is a minimum turning point.
$\left(2,4\right)$(2,4) is a minimum turning point.
$\left(-2,4\right)$(−2,4) is a maximum turning point.
Draw the graph below.
In the examples above we had to find the key features from an equation but we could be given pieces of information and be required to identify an appropriate graph or sketch a possible graph.
Sketch a possible graph that has the following properties:
Think: Carefully consider if the information given is about the function or its derivative.
The first clue tells us that the graph has stationary points at $x=2$x=2 and $x=1$x=1.
The second piece of information tells us the graph passes through $\left(2,0\right)$(2,0). This is also our stationary point from the first clue and an $x$x-intercept.
The third piece of information tells us that the graph also passes through $\left(1,1\right)$(1,1), which is our other stationary point.
Dot point $4$4, tells us that when $x=0$x=0, $y=-4$y=−4. That is the $y$y-intercept is $\left(0,-4\right)$(0,−4).
Dot point $5$5 tells us that between the two stationary points we have a decreasing function and elsewhere we have an increasing function.
Do: Plotting the points and indicating the known areas of increasing and decreasing, we obtain:
Joining the dots with a smooth curve and following the arrow trends we have a graph of a function with the given properties - this is not the only possibility, can you draw a few more?
Sketch the linear function that satisfies the following information
$f\left(0\right)=1$f(0)=1
$f'\left(2\right)=3$f′(2)=3
Plot the line on the graph
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 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): |
$f'\left(-1\right)=0$f′(−1)=0 |
$f'\left(4\right)=0$f′(4)=0 |
$f'\left(x\right)>0$f′(x)>0 for $x>4$x>4 |
$f'\left(x\right)<0$f′(x)<0 elsewhere |