Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Line (geometry) wikipedia , lookup
Dirac delta function wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
History of the function concept wikipedia , lookup
Function (mathematics) wikipedia , lookup
Proofs of Fermat's little theorem wikipedia , lookup
German tank problem wikipedia , lookup
Math 150 Must Show all Work! Quiz Name ____________________ 1. Sketch the graph x2 3 y= x3 x-intercepts: ( 3 ,0) Vertical Asymptote: x = 0 Horizontal Asymptote: y = 0 [2 x]( x 3 ) ( x 2 3)[3x 2 ] y’ = = (x3 )2 = x 4 9x 2 x6 9 x 2 (3 x)(3 x) x4 x4 the critical numbers are 3 and -3 ----- ++++ - ------ ++++ -3 0 y’ 3 By the First Derivative Test, we have a local minimum at x = -3 and a local maximum at x = 3. The local minimum value is -0.22 and the local maximum is 0.22. Local Minimum (-3, -0.22) and Local Maximum (3, 0.22) . 1 x 2. Describe the intervals on which f(x) = 1 x is concave up and concave down. Also list the points of inflection. 1 1 1 1 1 1 1 1 1 1 1 1 [ x 2 ](1 x 2 ) (1 x 2 )[ x 2 ] [ x 2 ](1 x 2 ) (1 x 2 )[ x 2 ] 2 2 2 2 f’(x) = = 1 1 (1 x 2 ) 2 (1 x 2 ) 2 (Multiply numerator and denominator to clear the smaller fractions) f’(x) = 1 2 1 2 (1 x ) (1 x ) 1 2 1 2 2 x (1 x ) 2 3 1 = 1 1 2 1 2 x 1 2 1 2 * (1 x ) x (1 x ) 2 1 1 1 1 1 1 f” (x) = [ x 2 ](1+ x 2 )-2 +(- x 2 )[-2( 1 x 2 ) 3 * x 2 ] = 2 2 2 2 12 2x 1 2x 2 = 3 2 1 2 1 2 3 2 1 2 1 2 1 1 x (1+ x )-2 +x-1 ( 1 x ) 3 = x (1+ x )-3 [(1+ x ) + 2 x ] = 2 2 Function undefined +++++++++++ 1 3x 1 2 3 1 2 1 2 x 2 (1 x 2 ) 3 Though not very exciting, we have shown that this function is concave up throughout its domain. y’’ 0 Here is the graph: 3. Describe the intervals on which f(x) = 2 cos 2 x x 2 for x in [0, ] is concave up and concave down. Also list the points of inflection. f’(x) = -4cosxsinx-2x = -2sin(2x) – 2x f’' (x) = -4cos(2x) – 2 PPI: cos(2x) = - 2 = 3 4 3 1 1 by letting = 2x cos( ) = ’ = 2 2 3 2k 2k 2x = (***Correction above***) 2 3 4 3 2k 2k x = 3 k 2 k 3 we get only 2 possible point of inflection x = and 2 that lies with in the interval [0, ] 3 3 sign chart to determine concavity : Concave down ------------ Concave up Conc. Down -------+++++ 3 2 3 We get two points of inflection 2 which is ( ,-0.60) & ( , -3.89) 3 3 y’’ 4. A window has the shape of a rectangle surmounted by a semicircle. If the total perimeter of the window is 20 feet, find the dimensions that will maximize the amount of light to enter. Let x represent the radius of the semicircle and y represent the height of the rectangular portion of the window: The function to be maximized is the area A = 2xy + 1 x2 2 Since the perimeter of the window is 20 feet we have 20 = 2y + 2x + x We can use this to rewrite y in terms of x: y = 10 – x - x (*) 2 Substituting (*) into the area function we obtain: 1 1 x ) + x2 = 20x -2x2 - x2 + x2 2 2 2 (Note that this is a quadratic function which means that the only critical number occurs at the vertex) A(x) = 2x(10 – x - A’(x) = 20 – 4x - 2 x + x = 20 - 4x - x by setting this equal to zero we obtain the critical number x = 20 4 2.80 since A’’ is easy to calculate, let’s use the Second Derivative Test to verify that this critical number results in a Maximum Area : A’’(x) = -4 - which shows that this function is always concave down and that the Local Maximum and Absolute Maximum Area occur at 20 2.80 x= 4 20 2.80 To find y we substitute this x value into (*) to obtain y = x = 4 Therefore, in order to maximize the amount of light to enter our window would have be constructed in the shape of a rectangle approximately 5.6 x 2.8 and the semicircular part would have a radius of approximately 2.8. 5. What is the maximum volume for a rectangular box (square base, no top) made from 12 square feet of cardboard? (Assume no wasted material) Let the dimensions of the base be x by x and the height be y. The function to be maximized is the volume: V = x2y Since we are constructing the box from 12 ft2 we have the equation: x2 + 4xy = 12 12 x 2 y= 4x V(x) = x2 * V’(x) = 3 - (**) 1 12 x 2 = 3x - x 3 4 4x 3 2 3 x = (4 x 2 ) 4 4 3 (2 x )( 2 x) 4 Which yields critical numbers x = 2 & x = -2 (not applicable) V’(x) = x=2 Again, let us use the Second Derivative Test 3 12 2 2 V’’(x) = - x V’’(2) = -3 and we have a Maximum at x = 2 & y = =1 2 4(2) The dimensions that yield the maximum volume are 2x2x1 and the Maximum Volume is 4 cubic feet!