# print barcode image c# Find all local maxima and minima of the function k(x) = x 3 3x 2 24x + 5 in Software Creator QR Code 2d barcode in Software Find all local maxima and minima of the function k(x) = x 3 3x 2 24x + 5

Find all local maxima and minima of the function k(x) = x 3 3x 2 24x + 5
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SOLUTION We begin by calculating the rst derivative: k (x) = 3x 2 6x 24 = 3(x + 2)(x 4) We notice that k vanishes only when x = 2 or x = 4 These are the only candidates for local maxima or minima The second derivative is k (x) = 6x 6 Now k (4) = 18 > 0, so x = 4 is the location of a local minimum Also k ( 2) = 18 < 0, so x = 2 is the location of a local maximum A glance at the graph of this function, as depicted in Fig 310, con rms our calculations EXAMPLE 37
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Find all local maxima and minima of the function g(x) = x + sin x
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SOLUTION First we calculate that g (x) = 1 + cos x Thus g vanishes at the points (2k + 1) for k = , 2, 1, 0, 1, 2, Now g (x) = sin x And g ((2k + 1) ) = 0 Thus the second derivative test is inconclusive Let us instead look at the rst derivative We notice that it is always 0 But, as we have already noticed, the rst derivative changes sign at a local maximum or minimum We conclude that none of the points (2k + 1) is either a maximum nor a minimum The graph in Fig 311 con rms this calculation You Try It: Find all local maxima and minima of the function g(x) = 2x 3 15x 2 + 24x + 6
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CHAPTER 3 Applications of the Derivative
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EXAMPLE 38
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A box is to be made from a sheet of cardboard that measures 12 12 The construction will be achieved by cutting a square from each corner of
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Fig 310
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Fig 311
CHAPTER 3 Applications of the Derivative
the sheet and then folding up the sides (see Fig 312) What is the box of greatest volume that can be constructed in this fashion
Fig 312
SOLUTION It is important in a problem of this kind to introduce a variable Let x be the side length of the squares that are to be cut from the sheet of cardboard Then the side length of the resulting box will be 12 2x (see Fig 313) Also the height of the box will be x As a result, the volume of the box will be V (x) = x (12 2x) (12 2x) = 144x 48x 2 + 4x 3 Our job is to maximize this function V
12 _ 2x
Fig 313
Now V (x) = 144 96x + 12x 2 We may solve the quadratic equation 144 96x + 12x 2 = 0 to nd the critical points for this problem Using the quadratic formula, we nd that x = 2 and x = 6 are the critical points Now V (x) = 96 + 24x Since V (2) = 48 < 0, we conclude that x = 2 is a local maximum for the problem In fact we can sketch a graph of V (x) using ideas from calculus and see that x = 2 is a global maximum We conclude that if squares of side 2 are cut from the sheet of cardboard then a box of maximum volume will result Observe in passing that if squares of side 6 are cut from the sheet then (there will be no cardboard left!) the resulting box will have zero volume This value for x gives a minimum for the problem
CHAPTER 3 Applications of the Derivative
EXAMPLE 39
A rectangular garden is to be constructed against the side of a garage The gardener has 100 feet of fencing, and will construct a three-sided fence; the side of the garage will form the fourth side What dimensions will give the garden of greatest area
SOLUTION Look at Fig 314 Let x denote the side of the garden that is perpendicular to the side of the garage Then the resulting garden has width x feet and length 100 2x feet The area of the garden is A(x) = x (100 2x) = 100x 2x 2