Finding Maxima and Minima Using Derivatives
Finding maxima and minima using derivatives involves using the first and second derivative tests to determine the critical points of a function and classify them as maxima, minima, or points of inflection. The first derivative test examines the sign changes around critical points, while the second derivative test analyzes concavity to confirm the nature of the extremum.
6 Key excerpts on "Finding Maxima and Minima Using Derivatives"
eBook - ePub- Gregory Hill, Mel Friedman(Authors)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...Recall that the first-derivative test for maxima and minima required identifying critical points and checking whether the first derivative changed signs on either side of the critical point. This requires knowledge about how the function is behaving around the critical point. The second-derivative test is a local test applied right at the critical point. Second Derivative Test for Local Extremes Let c be a point on the interior of the domain of a function, f. If f ′(c) = 0 and f ″(c) > 0, then f has a local minimum at x = c. If f ′(c) = 0 and f ″(c) < 0, then f has a local maximum at x = c. If f ′(c) = 0, f has a horizontal tangent at x = c. But if f′ (c) > 0, then f is concave up, so the graph must lie above the horizontal tangent at c. Therefore, f must have a local minimum at x = c. Similar logic can be applied for a local maximum. EXAMPLE 5.14 Use the second-derivative test to determine the x-coordinates of the local extreme values on f (x) = x 3 + 3 x 2 – 9 x + 2. SOLUTION f ′(x) = 3 x 2 + 6 x – 9, and f ″(x) = 6 x + 6. Solving f ′(x) = 0 gives 3(x 2 + 2 x – 3) or 3(x + 3) (x – 1) = 0. So the critical values are x = –3 and x = 1. f ″(–3) = –12, and f ″(l) = 12, so f ″(–3) < 0 and f ″(1) > 0. By the second-derivative test, f has a local maximum at x = – 3 and a local minimum at x = 1. CURVE SKETCHING Mastering the relationships between a function and its first and second derivatives takes a good amount of practice. One way to test your conceptual understanding of these relationships is to work graphically with the three functions. The goal is to be able to synthesize information in such a way as to move seamlessly between a function and its derivatives. Given information about first and second derivatives, you ought to be able to make a rough sketch of the function...
eBook - ePub- Rose G Davies(Author)
- 2020(Publication Date)
- CRC Press(Publisher)
...For example, at point A, point C, the derivative of the function is “0”, i.e. the change of the function at those points is “0”. Therefore, those points are extrema (which can also be called stagnation points). The values of the function at the extrema are extreme values. Extreme values It is important in actual practice to be able to identify where or when the extreme values of a function occur, because this information might be able to optimise the advantages, or minimise the risks, or indicate some limits. As indicated in previously, extrema occur at the derivative of a continuous function, “0”. The following shows how to identify the type of extrema of f (x), a function of x : If f′ (x) = 0, at x = x o (e.g., Point A in Figure 1.4, x o = 1.), the derivative before this point is “−”, i.e. f′ (x) < 0, and the function f (x) is decreasing; the derivative after this point is “+”, i.e. f′ (x) > 0, and the function f (x) is increasing, the value of the function at this point f (x o) is a local minimum; or if f′ (x) = 0,. and f″ (x) > 0, at x = x o, this f (x o) is a local minimum. If f′ (x) = 0, at x = x o (e.g., Point C in Figure 1.4, x o = 3.), and the derivative before this point is “+”, i.e. f′ (x) < 0, and the function f (x) is increasing; the derivative after this point is “−”, i.e. f′ (x) > 0, and the function f (x) is decreasing, the value of the function at this point f (x o) is a local maximum; or if f′ (x) = 0, and f″ (x) < 0, at x = x o,. this f (x o) is a local maximum. Please note that those extreme values are called local or relative maximum/minimum because the absolute maximum/minimum values can occur at other places. For some functions, the values of the functions can go to infinity. For example, in Figure 1.4, the value of the function f (x) can be greater than f (1) (Point A, the local maximum), if x ≫ 4 (after point D). Example 1.1 y = y (t) = t 3 − t 2 − 2 t + 1...
eBook - ePub- Adil H. Mouhammed(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...If it is positive, the function has a minimum value at the critical point(s). In contrast, if the second derivative is negative, the function has a maximum value at the critical point(s). Otherwise, the test fails and the function may have an inflection point. Example 1: Optimize the function y = 3x 2 - 12x. Solution: Find the first derivative and set it equal to zero. That is, f′(x) = dy/dx = 6x - 12 = 0 and solving for the critical point x c yields 6x = 12 and x c = 2. Now, find the second derivative for the function d 2 y/dx 2 = f″(x) = 6. Because the second derivative is positive, the function has a minimum value at the critical point x c = 2. Also, if we substitute various values of x in our function, the y’s values (or the values of the function) will be As can be seen, the function has a minimum value of (-12) at the critical point x c = 2. One should note from this example that we can find the minimum value of the function by using the first derivative only. As we know the critical value, which is x c = 2, we can take other values for x, such as x = 1 (which is less than the critical value) and x = 3 (which is higher than the critical value), and after substituting these values in the first derivative we can obtain f′(x) = - 6 and f′(x) = 6. Because the first derivative changes signs from negative to positive, the function must have a minimum value at the critical point. Example 2: Optimize the function y = f(x) = -2x 2 + 20x - 2. Solution: The first derivative is dy/dx = -4x + 20 = 0, and hence x c = 5. The second derivative is d 2 y/dx 2 = - 4. As the second derivative is negative, the function has a maximum value at the critical point x c = 5. One should note that by substituting various values of x in the given function, we can find the maximum value of the function. This is shown below: Similarly, by using the first derivative alone, one can determine the maximum value of the function...
eBook - ePub- Adil H. Mouhammed(Author)
- 2020(Publication Date)
- Routledge(Publisher)
...If it is positive, the function has a minimum value at the critical point(s). In contrast, if the second derivative is negative, the function has a maximum value at the critical point(s). Otherwise, the test fails and the function may have an inflection point. Example 1: Optimize the function y = 3x 2 - 12x. Solution: Find the first derivative and set it equal to zero. That is, f ′ (x) = dy/dx = 6 x − 12 = 0 and solving for the critical point x c yields 6x = 12 and x c = 2. Now, find the second derivative for the function d 2 y/dx 2 = f ? (x) = 6. Because the second derivative is positive, the function has a minimum value at the critical point x c = 2. Also, if we substitute various values of x in our function, the y’s values (or the values of the function) will be x − 2 − 1 0 1 2 3 y 36 15 0 − 9 − 12 − 9 As can be seen, the function has a minimum value of (−12) at the critical point x c = 2. One should note from this example that we can find the minimum value of the function by using the first derivative only. As we know the critical value, which is x c = 2, we can take other values for x, such as x = 1 (which is less than the critical value) and x = 3 (which is higher than the critical value), and after substituting these values in the first derivative we can obtain f ′(x) = - 6 and f ′(x) = 6. Because the first derivative changes signs from negative to positive, the function must have a minimum value at the critical point. Example 2: Optimize the function y = f(x) = −2x 2 + 20x - 2. Solution: The first derivative is dy/dx = − 4 x + 20 = 0, and hence x c = 5. The second derivative is d 2 y/dx 2 = − 4. As the second derivative is negative, the function has a maximum value at the critical point x c = 5. One should note that by substituting various values of x in the given function, we can find the maximum value of the function...
- J. Rosebush, Stu Schwartz(Authors)
- 2016(Publication Date)
- Research & Education Association(Publisher)
...Absolute extrema can occur at interior points or at endpoints of a function. The absolute extrema occur at points where the first derivative is either zero or nonexistent and the function is defined—or at endpoints of the function. 4. Critical points—these are points in the domain of a function at which the derivative is either equal to zero or does not exist. These are generally found when looking for max/min points. 5. Asymptotes—refer to chapter 4. 6. Graphs to illustrate curve sketching i. Point A is the absolute maximum point. Point C is a relative maximum point. Point B is a relative minimum point. Point D is the absolute minimum point. ii. Point J is the absolute maximum point, but not a relative maximum. Points F and H are relative maxima. Points G and I are relative minima. Point E is the absolute minimum point, but not a relative minimum. iii. The absolute minimum point of y = h (x) is (1, –3). At x = 1, h (x) is defined but its derivative does not exist. A point on a function consists of both the x and y values of the point, so when asked to find a minimum / maximum point, find both the x and y values of it. When asked to find the minimum / maximum value of a function, find only the y -value. iv. Find the absolute minimum point of f (x) = xe x. Step 1: Find f ′(x). f ′(x) = xe x + e x Step 2: Set f ′(x) = 0 and also check for points where f ′(x) does not. exist. f ′(x) = xe x + e x = 0 → e x (x + 1) = 0 → x = –1 (e x is positive for all values of x). This function has no points of nondifferentiability. Step 3: Check to see if there is a max or min at x = –1 by making a sign analysis chart for f ′(x)...
eBook - ePub- Arsen Melkumian(Author)
- 2012(Publication Date)
- Routledge(Publisher)
...5 Optimization of univariate functions Many problems in economics require identification of an optimal outcome. This task is achieved by maximizing or minimizing some economic entity. For example, a firm may seek to maximize profit π by choosing the optimal level of output Q. Since π is the difference between total revenue (TR) and total cost (TC) we can write where π (Q) is known as the objective function and Q is referred to as the choice variable. The objective function π (Q) is the function that we seek to optimize. Other problems in economics require optimizing across two or more choice variables: one such example would be to minimize a cost function with respect to capital and labor. Another example would be to maximize a profit function with respect to capital, labor and land. In this chapter we will confine ourselves to examining objective functions with one variable, otherwise known in calculus as univariate functions. Finding extreme values of univariate functions is one of the most widely known and applied areas of calculus. 5.1 Local and global extrema In mathematics, maxima and minima are known collectively as extrema. The following is a rigorous definition of local and global extrema. D EFINITION 5.1 Let f (x) be a function defined on an interval I and A = (a, f (a)) be a point on the graph of the function. Then f (x) has: (i) a global maximum at A if f (a) ≥ f (x) for all x ∈ I ; (ii) a global minimum at A if f (a) ≤ f (x) for all x ∈ I ; (iii) a local maximum at A if f (a) ≥ f (x) for all x in a neighborhood of a ; (iv) a local minimum at A if f (a) ≤ f (x) for all x in a neighborhood of a ; (v) a strict global...
