# positive second derivative

The graph of $$y=f(x)$$ is increasing and concave up on the interval $$(-2,0.5)\text{,}$$ which is connected to the fact that $$f''$$ is positive, and that $$f'$$ is positive and increasing on the same interval. 1 v The formula for the best quadratic approximation to a function f around the point x = a is. We read $$f''(x)$$ as $$f$$-double prime of $$x$$, or as the second derivative of $$f$$. They assume that all campaigns produce some increase in sales. In Figure1.87 below, we see two functions and a sequence of tangent lines to each. {\displaystyle {\frac {d^{2}y}{dx^{2}}}} Equivalently, the value of $$F''(t)$$ is negative throughout the given time interval. − What is the meaning of the function $$y = s'(t)$$ in the context of the given problem? At that point, the second derivative is 0, meaning that the test is inconclusive. Remember that to plot $$y = f'(x)\text{,}$$ it is helpful to first identify where $$f'(x) = 0\text{. The Second Derivative Test. The graph of \(y=g(x)$$ is decreasing and concave down on the (approximate) intervals $$(-5,-4)\text{,}$$ $$(-2.5,-2.2)\text{,}$$ $$(0,1.5)\text{,}$$ $$(2.5,3)\text{,}$$ and $$(5,5.5)\text{. That is. The second derivative may be used to determine local extrema of a function under certain conditions. d A differentiable function \(f$$ is increasing at a point or on an interval whenever its first derivative is positive, and decreasing whenever its first derivative is negative. Letting $$f$$ be a constant function shows that if the derivative can be zero, then the function need not be increasing. Fundamentally, we are beginning to think about how a particular curve bends, with the natural comparison being made to lines, which don't bend at all. For example, it can be tempting to say that $$-100$$ is bigger than $$-2\text{. A point where this occurs is called an inflection point. 0 For a certain function \(y = g(x)\text{,}$$ its derivative is given by the function pictured in Figure1.97. Decreasing? Recall that a function is increasing when its derivative is positive. u L Based on the data, on what approximate time intervals is the function $$y = h(t)$$ concave down? {\displaystyle d(d(u))} If, however, the function has a critical point for which f′(x) = 0 and the second derivative is negative at this point, then f has local maximum here. Use the given data and your work in (a) to estimate $$h''(5)\text{.}$$. , When the first derivative is positive, the function is always. Examples of functions that are everywhere concave up are $$y = x^2$$ and $$y = e^x\text{;}$$ examples of functions that are everywhere concave down are $$y = -x^2$$ and $$y = -e^x\text{.}$$. f }\), $$v$$ is increasing from $$0$$ ft/min to $$7000$$ ft/min approximately on the $$66$$-second intervals $$(0,1.1)\text{,}$$ $$(3,4.1)\text{,}$$ $$(6,7.1)\text{,}$$ and $$(9,10.1)\text{. u} The second derivative is the rate of change of the slope, or the curvature. ( Similarly, the righthand plot in Figure1.87 depicts a function that is concave down; in this case, we see that the tangent lines alway lie above the curve and that the slopes of the tangent lines are decreasing as we move from left to right.$$, Given a function $$f(x)$$ defined on the interval $$(a,b)\text{,}$$ we say that $$f$$ is increasing on $$(a,b)$$ provided that $$f(x)\lt f(y)$$ whenever $$a\lt x\lt y\lt b\text{. ) Clearly, the position of the vehicle at the point where the velocity reaches zero will be the maximum distance from the starting position – after this time, the velocity will become negative and the vehicle will reverse. ] is the second derivative of position (x) with respect to time. f ‘(x) goes from negative to positive at x = 1, the First Derivative Test tells us that there is a local minimum at x = 1. f (1) = 2 is the local minimum value. \(y = f(x)$$ such that $$f$$ is increasing on $$-3 \lt x \lt 3\text{,}$$ concave up on $$-3 \lt x \lt 0\text{,}$$ and concave down on $$0 \lt x \lt 3\text{. }$$, $$v$$ is increasing on the intervals $$(0,1.1)\text{,}$$ $$(3,4.1)\text{,}$$ $$(6,7.1)\text{,}$$ and $$(9,10.1)\text{. This notation is derived from the following formula: As the previous section notes, the standard Leibniz notation for the second derivative is In addition, for each, write several careful sentences in the spirit of those in Example1.88 that connect the behaviors of \(f\text{,}$$ $$f'\text{,}$$ and $$f''$$ (or of $$g\text{,}$$ $$g'\text{,}$$ and $$g''$$ in the case of the second function). Well it could still be a local maximum or a local minimum so let's use the first derivative test to find out. n Suppose that an object's velocity is given by the graph in the figure below. f , which is defined as:. Second Derivative. The expression on the right can be written as a difference quotient of difference quotients: This limit can be viewed as a continuous version of the second difference for sequences. }\)7Notice that in higher order derivatives the exponent occurs in what appear to be different locations in the numerator and denominator. [ n 2 If the curve is curving upwards, like a smile, there’s a positive second derivative; if it’s curving downwards like a frown, there's a negative second derivative; where the curve is a straight line, the second derivative is zero. For the position function $$s$$ with velocity $$v$$ and acceleration $$a\text{,}$$. The car is stopped during the third minute. The meaning of the derivative function still holds, so when we compute $$f''(x)\text{,}$$ this new function measures slopes of tangent lines to the curve $$y = f'(x)\text{,}$$ as well as the instantaneous rate of change of $$y = f'(x)\text{. Concave down. n }$$ This is because the curve $$y = s(t)$$ is concave up on these intervals, which corresponds to an increasing first derivative $$y =s'(t)\text{. We see that at point \(A$$ the value of $$f'(x)$$ is positive and relatively close to zero, and at that point the graph is rising slowly. What does it mean to say that a function is concave up or concave down? expression. }\) $$v$$ is constant on the intervals $$(2,3)\text{,}$$ $$(5,6)\text{,}$$ $$(8,9)\text{,}$$ and $$(11,12)\text{.}$$. Similarly, a function whose second derivative is negative will be concave down (also simply called concave), and its tangent lines will lie above the graph of the function. }\), $$y = g(x)$$ such that $$g$$ is increasing on $$-3 \lt x \lt 3\text{,}$$ concave down on $$-3 \lt x \lt 0\text{,}$$ and concave up on $$0 \lt x \lt 3\text{. Once stable companies can quickly find themselves sidelined. For instance, write something such as. Therefore, we can say that acceleration is positive whenever the velocity function is increasing. Look at the two tangent lines shown below in Figure1.77. There are three basic behaviors that an increasing function can demonstrate on an interval, as pictured below in Figure1.85: the function can increase more and more rapidly, it can increase at the same rate, or it can increase in a way that is slowing down. 2 Time \(t$$ is measured in minutes. }\) Then $$f$$ is concave up on $$(a,b)$$ if and only if $$f'$$ is increasing on $$(a,b)\text{;}$$ $$f$$ is concave down on $$(a,b)$$ if and only if $$f'$$ is decreasing on $$(a,b)\text{. Let \(f$$ be a function that is differentiable on an interval $$(a,b)\text{. . on an interval where \(v$$ is negative, $$s$$ is . . This pattern of starts and stops continues for a total of $$12$$ minutes, by which time the car has traveled a total of $$16,000$$ feet from its starting point. \end{equation*}, Negative numbers present an interesting tension between common language and mathematical language. If the function has a derivative, the sign of the derivative tells us whether the function is increasing or decreasing. x The second derivative is acceleration or how fast velocity changes. The reason the second derivative produces these results can be seen by way of a real-world analogy.  That is: When using Leibniz's notation for derivatives, the second derivative of a dependent variable y with respect to an independent variable x is written. Acceleration is defined to be the instantaneous rate of change of velocity, as the acceleration of an object measures the rate at which the velocity of the object is changing. j \newcommand{\gt}{>} d Put another (mathematical) way, the second derivative is positive. d }\) This is connected to the fact that $$g''$$ is negative, and that $$g'$$ is negative and decreasing on the same intervals. \newcommand{\lt}{<} We start with an investigation of a moving object. That is, although it is formed looking like a fraction of differentials, the fraction cannot be split apart into pieces, the terms cannot be cancelled, etc. This leaves only the rightmost curve in Figure1.86 to consider. In addition to asking whether a function is increasing or decreasing, it is also natural to inquire how a function is increasing or decreasing. Notice that we have to have the derivative strictly positive to conclude that the function is increasing. If the second derivative is positive at … ( ) \end{equation*} In Example1.88 that we just finished, we can replace $$s\text{,}$$ $$v\text{,}$$ and $$a$$ with an arbitrary function $$f$$ and its derivatives $$f'$$ and $$f''\text{,}$$ and essentially all the same observations hold. ∈ 2.4.2 Interpretation of the Second Derivative as a Rate of Change Remark 5. Example 6. 2. ) Velocity is increasing on $$0\lt t\lt 1.1\text{,}$$ $$3\lt t\lt 4.1\text{,}$$ $$6\lt t\lt 7.1\text{,}$$ and $$9\lt t\lt 10.1\text{;}$$ $$y = v(t)$$ is decreasing on $$1.1\lt t\lt 2\text{,}$$ $$4.1\lt t\lt 5\text{,}$$ $$7.1\lt t\lt 8\text{,}$$ and $$10.1\lt t\lt 11\text{. The second derivative is defined by applying the limit definition of the derivative to the first derivative. On these intervals, then, the velocity function is constant. What can we learn by taking the derivative of the derivative (the second derivative) of a function \(f\text{?}$$. On the leftmost curve in Figure1.85, draw a sequence of tangent lines to the curve. Suppose f ‘’ is continuous near c, 1. Since the second derivative is positive on either side of x = 0, then the concavity is up on both sides and x = 0 is not an inflection point (the concavity does not change). By taking the derivative of the derivative of a function f, we arrive at the second derivative, f ″. The second derivative of a function f can be used to determine the concavity of the graph of f. A function whose second derivative is positive will be concave up (also referred to as convex), meaning that the tangent line will lie below the graph of the function. 2 If the second derivative is positive, the rate of change is increasing; if the second derivative is negative, the rate of change is decreasing. {\displaystyle v_{j}(x)={\sqrt {\tfrac {2}{L}}}\sin \left({\tfrac {j\pi x}{L}}\right)} If a function's FIRST derivative is negative at a certain point, what does that tell you? The derivative of $$f$$ tells us not only whether the function $$f$$ is increasing or decreasing on an interval, but also how the function $$f$$ is increasing or decreasing. d Now the left-hand side gets the second derivative of y with respect to to x, is going to be equal to, well, we just use the power rule again, negative three times negative 12 is positive 36, times x to the, well, negative three minus one is negative four power, which we could also write as 36 over x to the fourth power. These most recent observations formally as the definitions of the original function is named \ ( '... The figure below provide data for \ ( a\ ) is positive ( +,!, it can be tempting to say that acceleration is given by derivative! Generalization of the terms concave up ( ) and slower rate back to a function is increasing v. Mean that the first derivative is increasing way to classify critical point and, in particular, to ﬁnd maxima... 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