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We know if \(f\) is differentiable at \(x=a\) then the graph of \(f\) is locally linear at \(x=a\) and \(f'(a)\) is the slope of the tangent line at the point \((a, f(a))\). In this section, we explore how local behavior near a point can be extrapolated to global behavior over an interval. First, we review Definition incdeccnstdefn from Section ConstantandLinearFunctions:
Definition.
Let \(f\) be a function defined on an interval \(I\). Then \(f\) is said to be:
increasing on \(I\) if, whenever \(a < b\), then \(f(a) < f(b)\). (i.e., as inputs increase, outputs increase.)
NOTE: The graph of an increasing function rises as one moves from left to right.
decreasing on \(I\) if, whenever \(a < b\), then \(f(a) > f(b)\). (i.e., as inputs increase, outputs decrease.)
NOTE: The graph of a decreasing function falls as one moves from left to right.
constant on \(I\) if \(f(a) = f(b)\) for all \(a\), \(b\) in \(I\). (i.e., outputs don’t change with inputs.)
NOTE: The graph of a function that is constant over an interval is a horizontal line.
Suppose a function satisfies \(f'(x) > 0\) for all \(x\) in an open interval \(I\). Then we know that not only is the graph of \(f\) locally linear on \(I\), but the slopes of all of the tangent lines are positive. This means that all of the tangent lines are increasing so it stands to reason that the function \(f\) is likewise increasing on \(I\). In other words, if a function is locally increasing on \(I\), then it is globally increasing on \(I\) as well.
We can apply the same reasoning above to situations where \(f'(x)<0\) for all \(x\) in \(I\), which implies \(f\) is decreasing on \(I\) or \(f'(x) = 0\) on \(I\), which implies \(f\) is constant on \(I\). In Calculus, you’ll learn this fact is a consequence of the Mean Value Theorem. In this text, we’ll just accept the following theorem is true and hope we’ve done enough hand-waving to deem it reasonable.
Thanks to GeoGebra, we can visualize Theorem firstderivatveandgraphs. In the first case below, we have two sections of a graph over which a function is increasing. By adjusting the slider, we can move the point across the graph, observe the corresponding tangent line, and see the value of the slope. Note that in both cases below, the slope of the tangent line, \(f'(x)\), is always positive.
\(f'(x) > 0\) for all \(x\) in \(I\).
Next, we have two sections of a graph over which a function is decreasing. Here, we observe the slopes of the tangent lines are always negative.
\(f'(x) < 0\) for all \(x\) in \(I\).
Last, but not least, we have the graph of a constant function. Here the slopes are always \(0\).
\(f'(x) = 0\) for all \(x\) in \(I\).
We can use Theorem firstderivatveandgraphs to help us determine the (open) intervals over which a function \(f\) is increasing, decreasing, and constant by making a sign diagram for the derivative \(f'\).
In order to avoid us having to go through the (somewhat lengthy) process of finding \(f'(x)\) using Definition derivativefcndefn, we’ll just use some properties of derivatives from Calculus behind the scenes and present you with both a function and its derivative. It’s time for an example.
We are given \(f'(x) = 3x^2-6x-9\). Solving \(f'(x) = 3x^2-6x-9 = 0\) gives \(3(x^2-2x-3) = 0\) or \(3(x-3)(x+1) = 0\). We get two solutions: \(x = -1\) and \(x = 3\) which divides the \(x\)-axis into three regions: \(x< -1\), \(-1<x<3\) and \(x>3\).
Next we select a test value in each of these three regions to determine the sign of \(f'(x)\). For the interval \(x<-1\), we select \(x = -3\): \(f'(-3) = 3(-3)^2-6(-3)-9 = (+)\). For \(-1<x<3\), we select \(x = 0\): \(f'(0) = 3(0)^2-6(0)-9 = (-)\). Finally, for \(x>3\), we select \(x = 4\): \(f'(4) = 3(4)^2-6(4)-9 = (+)\).
Next, we use Theorem firstderivatveandgraphs to interpret what the sign diagram for \(f'(x)\) means for the graph of \(y = f(x)\):
We find \(f\) is increasing on \((-\infty , -1)\) and again on \((3, \infty )\) while \(f\) is decreasing on \((-1,3)\). At the points \(x = -1\) and \(x=3\), we have \(f'(x) = 0\) so the graph of \(f\) is locally flat there.
Since \(f\) changes from increasing just to the left of \(x=-1\) to decreasing just to the right of \(x=-1\), it stands to reason that \(f\) has a local maximum at \(x=-1\). This is indeed the case and we find that the local maximum value is \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1)+5 = 10\).
Similarly, since \(f\) changes from decreasing just to the left of \(x=3\) to increasing just to the right of \(x=3\), \(f\) has a local minimum at \(x=3\). The local minimum value is \(f(3) = (3)^3 - 3(3)^2 - 9(3)+5 = -22\).
A quick check using desmos confirms our results. Note here the portions of the graph over which \(f\) is increasing are highlighted in orange while the interval over which \(f\) is decreasing is highlighted in blue.
We generalize our observations about local extrema in the following result.
In order to make a sign diagram for \(f'(x)\), we rewrite \(f'(x)\) as a single fraction:
Unlike the derivative in Example polyincdec, \(f'(x) = \frac {4x-4}{3x^{2/3}}\) is undefined when \(3x^{2/3} = 0\), that is, when \(x = 0\), so we need to record this on our sign diagram with the customary ‘‽.’
Next, we solve \(f'(x) = \frac {4x-4}{3x^{2/3}} = 0\) to get \(4x-4 = 0\) or \(x = 1\). The usual machinations produces the sign diagram for \(f'(x)\) below.
Using Theorem firstderivatveandgraphs, we get:
We get \(f\) is decreasing for \(x<0\) as well as from \(0 < x < 1\). Since \(0\) is in the domain of \(f\), we splice the two intervals together so \(f\) is decreasing from \((-\infty , 1)\). We see \(f\) is increasing from \((1, \infty )\).
We note that \(f\) satisfies the conditions of Theorem firstderivatvetest since \(f\) is continuous everywhere and \(f'\) exists for all \(x \neq 0\). Since \(f\) changes from decreasing just to the left of \(x=1\) to increasing just to the right of \(x=1\), the graph of \(f\) has a local minimum at \(x=1\). The local minimum value is \(f(1) = (1)^{4/3} - 4(1)^{1/3} = -3\).
What is happening at \(x = 0\)? Since \(f'(x)\) doesn’t change sign on either side of \(0\), the graph of \(f\) doesn’t have a local extremum there. The sign diagram indicates \(f\) is decreasing through that point. A quick check using desmos reveals ‘unsual steepness’ at \(x = 0\), a phenomenon which is called a vertical tangent. This means the function locally resembles a vertical line.
We confirm our results with desmos. Once again, the portion of the graph over which \(f\) is decreasing is highlighted in blue while the portion of the graph over which \(f\) is increasing is highlighted in orange.
In section Section ??, we introduced the notion of concavity. In that section, we described curves as being concave up over an interval if it resembles a portion of a ‘\(\smile \)’ shape and concave down over an interval if resembles part of a ‘\(\frown \)’ shape. Now that we’ve had some exposure to Calculus, we can more precisely define these notions.
If we take the time to study a generic concave up curve, the ‘\(\smile \)’ shape can be divided into a decreasing and increasing arc. Using the GeoGebra interactive below, we notice that as we move the slider from left to right on the decreasing portion of the curve, the slopes are increasing towards \(0\).
Moving the slider on the second portion of the graph, we see the slopes are increasing from \(0\) as we move fron left to right.
In both of these cases, the slopes of the tangent line are increasing.
Likewise, we can dissect a generic ‘\(\frown \)’ shape curve into an increasing and decreasing arc. Here, as we move from left to right on the increasing portion, the slopes are decreasing to \(0\),
Moving the slider on the second portion of the graph from left to right, we find the slopes continuing to decrease from \(0\).
Here, the slopes of the tangent line are decreasing.
We know from Theorem 1 that the derivative of a function can tell us where that function is increasing and decreasing. Since the function which gives us the slopes of tangent lines is the derivative, \(f'(x)\), we could use the derivative of \(f'(x)\) to determine where the slopes of the tangent lines were increasing and decreasing. This leads us to define the second derivative, \(f''(x)\) as the derivative of \(f'(x)\).
We present the following theorem without proof, but hopefully sufficiently motivated.
Solving \(f''(x) = 6x-6 = 0\) gives \(x=1\). We find \(f''(0) = 6(0)-6 = (-)\) and \(f''(2) = 6(2) - 6= (+)\). This gives us our sign diagram for \(f''(x)\) below.
Using Theroem 3, we get the following:
We find \(f\) is concave down on \((-\infty , 1)\) and concave up on \((1, \infty )\).
At \(x=1\), the concavity changes. We find \(f(1) = (1)^3 - 3(1)^2 - 9(1)+5 = -6\) and we call the point \((1, -6)\) an inflection point. In this case since the concavity changes from concave down to concave up, the point \((1,-6)\) is the point on the graph of \(y=f(x)\) were the slopes stop decreasing and start to increase.
A quick check using desmos confirms our results. Here the portion of the graph which is concave down is highlighted in blue whereas the portion of the graphh which is concave up is highlighted in orange.
Note that we can use concavity to help us distinguish local extrema.
For the function above, both \(f'(-1) = 0\) and \(f'(3) = 0\). Note that \(f''(-1) < 0\) which means \(f\) is concave down there. This forces \(f\) to have a local maximum at \((-1,6)\). Likewise, \(f''(3) > 0\) which means \(f\) is concave up there. This forces \(f\) to have a local minimum at \((3,-22)\). We generalize this observation below.
If \(f''(c) = 0\) then the test is inconclusive. \(f\) may or may not have a local extremum at \(x=c\).
(In this case, we would appeal to the first derivative test.)
As in Example 2, our first step is to rewrite \(f''(x) = \frac {4}{9} x^{-2/3} + \frac {8}{9} x^{-5/3}\) as a single fraction:
We see \(f''(x) = \frac {4x+8}{9x^{5/3}} \) is undefined when \(9x^{5/3}= 0\), that is, when \(x = 0\).
Solving \(f''(x) = \frac {4x+8}{9x^{5/3}} = 0\) gives \(4x+8 = 0\) so \(x = -2\).
Going through the usual routine, we obtain our sign diagram for \(f''(x)\).
Using Theorem 3 we get:
We see \(f\) is concave up on \((-\infty , -2)\) and again from \((0, \infty )\). \(f\) is concave down on \((-2,0)\).
Since \(f\) changes concavity at both \(x=-2\) and \(x=0\), we have inflection point at both of these values.
We find: \(f(-2) = (-2)^{4/3} - 4(-2)^{1/3} = 2 (2)^{1/3} + 4 (2)^{1/3} = 6 (2)^{1/3}\). So \(\left (-2, 6 (2)^{1/3} \right )\) is one inflection point. When \(x = 0\), \(f(0) = (0)^{4/3} - 4(0)^{1/3} = 0\), so \((0,0)\) is the other inflection point.
Our graph below highlights the concave up portions of the graph in orange and the concave down portion of the graph in blue. While it’s not apparent that the graph of \(y=f(x)\) is concave up for \(x<-2\), we invite the reader to zoom out until that characteristic is more pronounced.
Our last example offers a twist on these sorts of curve-sketching problems.
Use the graph of \(y=f'(x)\) to determine the open intervals where \(f\) is increasing and decreasing.
Find the \(x\)-coordinates of the local extrema.
List the open intervals over which the graph of \(f\) is concave up and concave down.
Find the \(x\)-coordinates of the inflection points.
Recall from algebra, the solutions to \(f'(x) < 0\) are the \(x\)-values where the graph of \(y = f'(x)\) is below the \(x\)-axis. This happens on the intervals \((-\infty , 0)\) and \((4, \infty )\), so this means \(f\) is decreasing here.
Likewise, the solutions to \(f'(x) > 0\) are the \(x\)-values where \(y=f'(x)\) is above the \(x\)-axis. This happens on the interval \((0,4)\), so \(f\) is increasing here.
Since \(f\) goes from decreasing to the left of \(x=0\) to increasing to the right of \(x=0\), \(f\) has a local minimum at \(x=0\). Since \(f\) goes from increasing to the left of \(x=4\) to decreasing to the right of \(x=4\), \(f\) has a local maximum at \(x=4\).
Since \(f''(x)\) is the derivative of \(f'(x)\), we know \(f''(x) > 0\) on \((-\infty , 2)\) since \(f'(x)\) is increasing there. We see \(f''(2) = 0\) since \(f'(x)\) is locally flat at \((2,4)\). Lastly, we see \(f''(x) < 0\) on \((2, \infty )\) since \(f'(x)\) is decreasing there. We put all this together in a sign diagram below.
Hence, we know by Theorem 3:
We have \(f\) is concave up on \((-\infty , 2)\) and concave down on \((2, \infty )\).
Since \(f\) changes concavity at \(x=2\), there is an inflection point there.
A plausible graph of \(y = f(x)\) is below. Note that we cannot determine any \(y\)-coordinates (why not?)