Figure $$\PageIndex{7}$$: Number line for $$f$$ in Example $$\PageIndex{2}$$. We were careful before to use terminology "possible point of inflection'' since we needed to check to see if the concavity changed. A graph is concave up where its second derivative is positive and concave down where its second derivative is negative. In other words, the graph of f is concave up. We start by finding $$f'(x)=3x^2-3$$ and $$f''(x)=6x$$. Interval 2, $$(-1,0)$$: For any number $$c$$ in this interval, the term $$2c$$ in the numerator will be negative, the term $$(c^2+3)$$ in the numerator will be positive, and the term $$(c^2-1)^3$$ in the denominator will be negative. We utilize this concept in the next example. Let $$f$$ be differentiable on an interval $$I$$. Pick any $$c<0$$; $$f''(c)<0$$ so $$f$$ is concave down on $$(-\infty,0)$$. Notice how $$f$$ is concave up whenever $$f''$$ is positive, and concave down when $$f''$$ is negative. The second derivative of a function f can be used to determine the concavity of the graph of f. A function whose second derivative is positive will be concave up (also referred to as convex), meaning that the tangent line will lie below the graph of the function. Figure $$\PageIndex{11}$$: A graph of $$f(x) = x^4$$. Reading: Second Derivative and Concavity Graphically, a function is concave up if its graph is curved with the opening upward (figure 1a). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Let $$c$$ be a critical value of $$f$$ where $$f''(c)$$ is defined. Figure $$\PageIndex{6}$$: A graph of $$f(x)$$ used in Example$$\PageIndex{1}$$, Example $$\PageIndex{2}$$: Finding intervals of concave up/down, inflection points. Let $$f(x)=x/(x^2-1)$$. ", "When he saw the light turn yellow, he floored it. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Solving $$f''x)=0$$ reduces to solving $$2x(x^2+3)=0$$; we find $$x=0$$. (1 vote) Ï 2-XL Ï Reading: Second Derivative and Concavity. If the function is decreasing and concave down, then the rate of decrease is decreasing. THeorem $$\PageIndex{2}$$: Points of Inflection. Find the point at which sales are decreasing at their greatest rate. Over the first two years, sales are decreasing. We find $$f'(x)=-100/x^2+1$$ and $$f''(x) = 200/x^3.$$ We set $$f'(x)=0$$ and solve for $$x$$ to find the critical values (note that f'\ is not defined at $$x=0$$, but neither is $$f$$ so this is not a critical value.) Since the concavity changes at $$x=0$$, the point $$(0,1)$$ is an inflection point. We now apply the same technique to $$f'$$ itself, and learn what this tells us about $$f$$. Evaluating $$f''(-10)=-0.1<0$$, determining a relative maximum at $$x=-10$$. Find the inflection points of $$f$$ and the intervals on which it is concave up/down. Thus the derivative is increasing! This leads to the following theorem. Figure 1 The important $$x$$-values at which concavity might switch are $$x=-1$$, $$x=0$$ and $$x=1$$, which split the number line into four intervals as shown in Figure $$\PageIndex{7}$$. Figure $$\PageIndex{3}$$: Demonstrating the 4 ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives. A function is concave down if its graph lies below its tangent lines. This possible inflection point divides the real line into two intervals, $$(-\infty,0)$$ and $$(0,\infty)$$. "Wall Street reacted to the latest report that the rate of inflation is slowing down. If the second derivative of a function f(x) is defined on an interval (a,b) and f ''(x) > 0 on this interval, then the derivative of the derivative is positive. Note: We often state that "$$f$$ is concave up" instead of "the graph of $$f$$ is concave up" for simplicity. So the point $$(0,1)$$ is the only possible point of inflection. The following theorem officially states something that is intuitive: if a critical value occurs in a region where a function $$f$$ is concave up, then that critical value must correspond to a â¦ Similarly if the second derivative is negative, the graph is concave down. Sometimes, rather than using the first derivative test for extrema, the second derivative test can also help you to identify extrema. Thus the concavity changes where the second derivative is zero or undefined. Notice how the tangent line on the left is steep, upward, corresponding to a large value of $$f'$$. This is the point at which things first start looking up for the company. We have been learning how the first and second derivatives of a function relate information about the graph of that function. That means as one looks at a concave up graph from left to right, the slopes of the tangent lines will be increasing. In the numerator, the $$(c^2+3)$$ will be positive and the $$2c$$ term will be negative. Since the domain of $$f$$ is the union of three intervals, it makes sense that the concavity of $$f$$ could switch across intervals. We want to maximize the rate of decrease, which is to say, we want to find where $$S'$$ has a minimum. If a function is increasing and concave down, then its rate of increase is slowing; it is "leveling off." If the second derivative of a function f (x) is defined on an interval (a,b) and f '' (x) > 0 on this interval, then the derivative of the derivative is positive. We essentially repeat the above paragraphs with slight variation. The graph of $$f$$ is concave up if $$f''>0$$ on $$I$$, and is concave down if $$f''<0$$ on $$I$$. Moreover, if $$f(x)=1/x^2$$, then $$f$$ has a vertical asymptote at 0, but there is no change in concavity at 0. On the right, the tangent line is steep, downward, corresponding to a small value of $$f'$$. If $$f''(c)>0$$, then the graph is concave up at a critical point $$c$$ and $$f'$$ itself is growing. The second derivative $$f''(x)$$ tells us the rate at which the derivative changes. To show that the graphs above do in fact have concavity claimed above here is the graph again (blown up a little to make things clearer). The derivative measures the rate of change of $$f$$; maximizing $$f'$$ means finding the where $$f$$ is increasing the most -- where $$f$$ has the steepest tangent line. Figure $$\PageIndex{12}$$: Demonstrating the fact that relative maxima occur when the graph is concave down and relatve minima occur when the graph is concave up. Because f(x) is a polynomial function, its domain is all real numbers. Thus the numerator is negative and $$f''(c)$$ is negative. Our study of "nice" functions continues. In general, concavity can change only where either the second derivative is 0, where there is a vertical asymptote, or (rare in practice) where the second derivative is undefined. If $$f'$$ is constant then the graph of $$f$$ is said to have no concavity. This is both the inflection point and the point of maximum decrease. ", "As the immunization program took hold, the rate of new infections decreased dramatically. This leads us to a definition. A function whose second derivative is being discussed. Second Derivative. The intervals where concave up/down are also indicated. There is only one point of inflection, $$(0,0)$$, as $$f$$ is not defined at $$x=\pm 1$$. The Second Derivative Test The Second Derivative Test relates the concepts of critical points, extreme values, and concavity to give a very useful tool for determining whether a critical point on the graph of a function is a relative minimum or maximum. That is, sales are decreasing at the fastest rate at $$t\approx 1.16$$. The function is decreasing at a faster and faster rate. Similarly, a function is concave down if â¦ Figure $$\PageIndex{1}$$: A function $$f$$ with a concave up graph. The graph of $$f$$ is concave up on $$I$$ if $$f'$$ is increasing. Time saving links below. If the second derivative is positive at a point, the graph is bending upwards at that point. So, as you can see, in the two upper graphs all of the tangent lines sketched in are all below the graph of the function and these are concave up. To find the inflection points, we use Theorem $$\PageIndex{2}$$ and find where $$f''(x)=0$$ or where $$f''$$ is undefined. The second derivative is evaluated at each critical point. That means as one looks at a concave down graph from left to right, the slopes of the tangent lines will be decreasing. Figure 1. If knowing where a graph is concave up/down is important, it makes sense that the places where the graph changes from one to the other is also important. Example $$\PageIndex{4}$$: Using the Second Derivative Test. A similar statement can be made for minimizing $$f'$$; it corresponds to where $$f$$ has the steepest negatively--sloped tangent line. View Concavity_and_2nd_derivative_test.ppt from MATH NYA 201-NYA-05 at Dawson College. Interval 3, $$(0,1)$$: Any number $$c$$ in this interval will be positive and "small." Thus the numerator is positive while the denominator is negative. The graph is concave down when the second derivative is negative and concave up when the second derivative is positive. Legal. If second derivative does this, then it meets the conditions for an inflection point, meaning we are now dealing with 2 different concavities. Missed the LibreFest? Similarly, if f ''(x) < 0 on (a,b), then the graph is concave down. Pick any $$c>0$$; $$f''(c)>0$$ so $$f$$ is concave up on $$(0,\infty)$$. Notice how the slopes of the tangent lines, when looking from left to right, are increasing. Subsection 3.6.3 Second Derivative â Concavity. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "article:topic", "second derivative test", "Concavity", "Second Derivative", "inflection point", "authorname:apex", "showtoc:no", "license:ccbync" ], $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$. The second derivative test for concavity states that: If the 2nd derivative is greater than zero, then the graph of the function is concave up. Note that we need to compute and analyze the second derivative to understand concavity, so we may as well try to use the second derivative test for maxima and minima. Test for Concavity â¢Let f be a function whose second derivative exists on an open interval I. Find the inflection points of $$f$$ and the intervals on which it is concave up/down. If the concavity of $$f$$ changes at a point $$(c,f(c))$$, then $$f'$$ is changing from increasing to decreasing (or, decreasing to increasing) at $$x=c$$. Concavity Using Derivatives You can easily find whether a function is concave up or down in an interval based on the sign of the second derivative of the function. The Second Derivative Test The first derivative of a function gave us a test to find if a critical value corresponded to a relative maximum, minimum, or neither. This section explores how knowing information about $$f''$$ gives information about $$f$$. Consider Figure $$\PageIndex{1}$$, where a concave up graph is shown along with some tangent lines. ". Figure $$\PageIndex{3}$$: Demonstrating the 4 ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives. In the lower two graphs all the tangent lines are above the graph of the function and these are concave down. Hence its derivative, i.e., the second derivative, does not change sign. We find $$S'(t)=4t^3-16t$$ and $$S''(t)=12t^2-16$$. Pre Algebra. What does a "relative maximum of $$f'$$" mean? The denominator of $$f''(x)$$ will be positive. We begin with a definition, then explore its meaning. Note: Geometrically speaking, a function is concave up if its graph lies above its tangent lines. The second derivative gives us another way to test if a critical point is a local maximum or minimum. But concavity doesn't \emph{have} to change at these places. When $$f''<0$$, $$f'$$ is decreasing. Similarly, a function is concave down if its graph opens downward (Figure 1b). This means the function goes from decreasing to increasing, indicating a local minimum at $$c$$. The inflection points in this case are . A positive second derivative means that section is concave up, while a negative second derivative means concave down. The graph of $$f$$ is concave down on $$I$$ if $$f'$$ is decreasing. We conclude that $$f$$ is concave up on $$(-1,0)\cup(1,\infty)$$ and concave down on $$(-\infty,-1)\cup(0,1)$$. Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics. Not every critical point corresponds to a relative extrema; $$f(x)=x^3$$ has a critical point at $$(0,0)$$ but no relative maximum or minimum. We also note that $$f$$ itself is not defined at $$x=\pm1$$, having a domain of $$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$$. It is admittedly terrible, but it works. This calculus video tutorial provides a basic introduction into concavity and inflection points. THeorem $$\PageIndex{1}$$: Test for Concavity. It is evident that $$f''(c)>0$$, so we conclude that $$f$$ is concave up on $$(1,\infty)$$. Now consider a function which is concave down. Evaluating $$f''$$ at $$x=10$$ gives $$0.1>0$$, so there is a local minimum at $$x=10$$. THeorem $$\PageIndex{3}$$: The Second Derivative Test. It this example, the possible point of inflection $$(0,0)$$ is not a point of inflection. Thus $$f''(c)>0$$ and $$f$$ is concave up on this interval. We find the critical values are $$x=\pm 10$$. When $$S'(t)<0$$, sales are decreasing; note how at $$t\approx 1.16$$, $$S'(t)$$ is minimized. The second derivative shows the concavity of a function, which is the curvature of a function. Thus $$f''(c)<0$$ and $$f$$ is concave down on this interval. Our work is confirmed by the graph of $$f$$ in Figure $$\PageIndex{8}$$. Interval 4, $$(1,\infty)$$: Choose a large value for $$c$$. The following theorem officially states something that is intuitive: if a critical value occurs in a region where a function $$f$$ is concave up, then that critical value must correspond to a â¦ An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The sign of the second derivative gives us information about its concavity. If the graph of a function is linear on some interval in its domain, its second derivative will be zero, and it is said to have no concavity on that interval. Notice how $$f$$ is concave down precisely when $$f''(x)<0$$ and concave up when $$f''(x)>0$$. Free companion worksheets. Concavity and Second Derivatives. A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. On the interval of $$(1.16,2)$$, $$S$$ is decreasing but concave up, so the decline in sales is "leveling off.". Conversely, if the graph is concave up or down, then the derivative is monotonic. Note that we need to compute and analyze the second derivative to understand concavity, which can help us to identify whether critical points correspond to maxima or minima. The second derivative gives us another way to test if a critical point is a local maximum or minimum. If "( )>0 for all x in I, then the graph of f is concave upward on I. The key to studying $$f'$$ is to consider its derivative, namely $$f''$$, which is the second derivative of $$f$$. If for some reason this fails we can then try one of the other tests. We can apply the results of the previous section and to find intervals on which a graph is concave up or down. Our definition of concave up and concave down is given in terms of when the first derivative is increasing or decreasing. In the next section we combine all of this information to produce accurate sketches of functions. When $$f''>0$$, $$f'$$ is increasing. Find the domain of . Such a point is called a point of inflection. We use a process similar to the one used in the previous section to determine increasing/decreasing. Notice how the slopes of the tangent lines, when looking from left to right, are decreasing. Figure $$\PageIndex{13}$$: A graph of $$f(x)$$ in Example $$\PageIndex{4}$$. To find the possible points of inflection, we seek to find where $$f''(x)=0$$ and where $$f''$$ is not defined. A graph of $$S(t)$$ and $$S'(t)$$ is given in Figure $$\PageIndex{10}$$. Figure $$\PageIndex{4}$$: A graph of a function with its inflection points marked. That is, we recognize that $$f'$$ is increasing when $$f''>0$$, etc. Keep in mind that all we are concerned with is the sign of $$f''$$ on the interval. We find that $$f''$$ is not defined when $$x=\pm 1$$, for then the denominator of $$f''$$ is 0. Using the Quotient Rule and simplifying, we find, $f'(x)=\frac{-(1+x^2)}{(x^2-1)^2} \quad \text{and}\quad f''(x) = \frac{2x(x^2+3)}{(x^2-1)^3}.$. These results are confirmed in Figure $$\PageIndex{13}$$. Algebra. Clearly $$f$$ is always concave up, despite the fact that $$f''(x) = 0$$ when $$x=0$$. Replace the variable with in the expression . Perhaps the easiest way to understand how to interpret the sign of the second derivative is to think about what it implies about the slope of â¦ Notice how the tangent line on the left is steep, downward, corresponding to a small value of $$f'$$. The Second Derivative Test relates to the First Derivative Test in the following way. Let $$f$$ be twice differentiable on an interval $$I$$. And where the concavity switches from up to down or down to up (like at A and B), you have an inflection point, and the second derivative there will (usually) be zero. Exercises 5.4. We find $$f''$$ is always defined, and is 0 only when $$x=0$$. We technically cannot say that $$f$$ has a point of inflection at $$x=\pm1$$ as they are not part of the domain, but we must still consider these $$x$$-values to be important and will include them in our number line. The canonical example of $$f''(x)=0$$ without concavity changing is $$f(x)=x^4$$. Figure $$\PageIndex{4}$$ shows a graph of a function with inflection points labeled. Thus the derivative is increasing! Since $$f'(c)=0$$ and $$f'$$ is growing at $$c$$, then it must go from negative to positive at $$c$$. The graph of a function $$f$$ is concave up when $$f'$$ is increasing. 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