Consider the two curves pictured below. Compare and contrast them. What characteristics of these functions are the same? What characteristics of these functions are different?
Make a conjecture about the rate of change of both \(f'\) and \(g'\text{.}\) We’ll call these things second derivative functions, \(f''(x)\) and \(g''(x)\text{.}\)
We say that a function is concave up on an interval \((a,b)\) if \(f'(x)\) is increasing on the interval. If \(f'(x)\) is decreasing on the interval, then we say that \(f(x)\) is concave down.
We say that a point \((c, f(c))\) is an inflection point if it is a point at which \(f\) changes concavity (from concave up to concave down or vice versa).
Note that when we think about a function being concave up or down on some interval, we can think about this in different ways. Curvature can sometimes be hard to recognize visually, but one of the things we can see from the visual above is the interaction between the tangent line and the curve:
If the function is concave up on some interval, then the tangent line sits below the function at every point on that interval. The function curves upward away from the tangent line. Sometimes people will say that the curvature is concave “up, like a cup.”
If the function is concave down on some interval, then the tangent line sits above the function at every point on that interval. The function curves downward away from the tangent line. Sometimes people will say that the curvature is concave “down, like a frown.”
So we have some visual ways of thinking about these different types of curvature. Annoyingly, though, it is still relatively difficult to pinpoint the exact (or even close) location of an inflection point. We might be able to pretty easily point at the locations of local maximums and local minimums on a graph of a function, but it can be hard to see the exact point at which a graph of a function changes from concave up to down or vice versa. We’ll focus on finding them algebraically first, but then we’ll think a bit more about the graphs of functions later.
From there, we can build a little sign chart, where we split up the \(x\)-values based on the domain of \(f\) and the critical numbers of \(f'\text{.}\) Then we can attach each interval of \(x\)-values with a sign of the second derivative, \(f''\text{,}\) on that interval.
We’ll interpret these signs. For any intervals where \(f''(x)\gt 0\text{,}\) we know that \(f'(x)\) must be increasing and so \(f(x)\) is concave up. Similarly, for any intervals where \(f''(x)\lt 0\text{,}\) we know that \(f'(x)\) must be decreasing and so \(f(x)\) is concave down.
While we have this first derivative, we could find the critical points of \(f(x)\) and then classify those critical points using the First Derivative Test.
For our goal of finding the intervals where \(f(x)\) is concave up and concave down, and then finding the inflection points of \(f\text{,}\) let’s move on.
Now solve this for \(u\) using the quadratic formula. Note that in the end, the two values you get will be possibilities for \(u=x^2\text{.}\) Make sure you get your answer to be in terms of \(x\text{!}\)
You have two critical points of \(f'(x)\) (let’s just call them \(x_1\) and \(x_2\)). These are possible inflection points of \(f(x)\text{,}\) but we need to check to see if the concavity changes at these points.
Move the point on any graph and make sure the statements about signs, directions, and concavity match what you found! You should notice that signs of the first and second derivative change at the local maximums/minimums and the inflection points that we found.
We can notice that, by the definition, any inflection point is a point at which \(f(x)\) changes concavity, and so is a point where \(f'(x)\) changes direction. That means we are looking at local maximums or local minimums of \(f'(x)\) (as long as they’re not at the end points of some domain)! Similarly, these are points at which \(f''(x)\) changes sign, and so we are thinking about the \(x\)-intercepts of these second derivatives (or other kinds of locations where the second derivative could change signs).
Consider the function \(f(x) = \ln(x^2+1)\text{.}\) Find the intervals where \(f\) is concave up, the intervals where it is concave down, and then find the locations of any inflection points.
Consider the function \(g(x) = e^{-x^2}\text{.}\) Find the intervals where \(g\) is concave up, the intervals where it is concave down, and then find the locations of any inflection points.
At this point, we have three different functions that we are juggling: a function \(f(x)\) (or whatever name we give it), the first derivative \(f'(x)\text{,}\) and the second derivative \(f''(x)\text{.}\) We’ll want to keep in mind the role of each of these.
\(f(x)\) tells us the height, the \(y\)-value, of the function at some point.
Notice that we could extend this table and think about any triplet of functions/derivatives in a row: we just need to think about what is positive/negative, what is increasing/decreasing, and what is concave up/down. If we wanted, we could try to define some adjective to describe what is happening to a function when \(f'(x)\) is concave up/down, but let’s not. It’s hard enough to visualize concavity, and it will be difficult to visualize rates of change of the concavity.
Theorem4.3.3.Second Derivative Test for Local Maximums or Local Minimums.
If \((c,f(c))\) is a critical point of \(f\) with \(f'(c)=0\text{,}\) then we can use the value of the second derivative at \(x=c\) to classify the critical point:
If \(f''(c)\gt 0\text{,}\) then \(f\) is concave up at and around \(x=c\text{,}\) and so the function has a local minimum at \((c,f(c))\text{.}\)
Find any critical points of the following functions. For each, use the Second Derivative Test to classify the critical point. If the Second Derivative Test fails (we get that the second derivative evaluated at the critical point is 0, and so is inconclusive), then classify the critical point in some other way.
Let’s leave this here, with a few questions for you to think about:
When, for you, do you think it would be reasonable to use the Second Derivative Test instead of the First Derivative Test? What goes into making this decision?
When, for you, do you think it would be reasonable to use the First Derivative Test instead of the Second Derivative Test? What goes into making this decision?
Consider a function \(f(x)\) that is continuous on \((-\infty, \infty)\) with \(f''(x) \gt 0\) on \((-\infty, 5)\) and \(f''(x)\lt 0\) on \((5,\infty)\text{.}\) What do we know about \(f'(x)\) on these intervals? What do we know about \(f(x)\) on these intervals?
Consider a function \(g(x)\) that is continuous on \((-\infty, \infty)\) that is concave down on \((-\infty, -2)\text{,}\) concave up on \((-2,4)\text{,}\) and concave down on \((4,\infty)\text{.}\) Where are the inflection points of \(g(x)\text{?}\) What is happening to \(g'(x)\) at these points?
Explain (with words and/or sketches) why the Second Derivative Test requires \(f'(a)=0\) for the second derivative to classify the critical number \(a\) as a local maximum or minimum. Why can’t the second derivative classify a critical point where \(f'(a)\) does not exist?
For each of the following functions, use the second derivative to find the intervals where the function is concave up or concave down, and find the location of any inflection points.
Find critical points of the following functions, and apply the Second Derivative Test to classify those critical points as local maximums/minimums. If the test fails, explain why.
Sketch a graph of a function \(f(x)\) and has the following characteristics. Label the intervals where \(f\) is concave up/down, and find the locations of any local inflection points.
Explain whether \(x=a\) is the location of a local max of \(f(x)\text{,}\) a local min of \(f(x)\text{,}\) or neither. Reference both the First and Second Derivative Test.
If there is some \(x\)-value, (call it \(x_0\)) with \(a\lt x_0\lt b\) where we know that \(g'(x_0)=0\text{,}\) then what else can we say about this point on the graph of \(g(x)\text{?}\)
