Math 142 Lecture Notes. Section 5.2 Second Derivative and Graphs
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1 Math 142 Lecture Notes Section 5.2 Second Derivative and Graphs A) Derivatives of Higher Order Let f be a function. The derivative of f denoted by f ' is a another function and therefore we can find its derivative denoted by f ''. Following with this process we can obtain derivatives of higher order. First Derivative Second Derivative n th derivative f ' (x) f''' (x) f (n) (x) df dx d 2 f dx 2 d n f dx n Find the first three derivatives of the function f x =4x 3 5x 2 20x 100 f ' x =12x 2 10x 20 => f '' x =24x 10 => f ''' x =24 B) Concavity Consider the graphs of the functions f x =x 2 and g x = x f(x) g(x)
2 They are increasing functions but with different shapes. However, for the function f its derivative f ' is increasing all the time and for the function g its derivative is decreasing all the time. Thus, the curvature is different in each case and that information is given by f ''. Summarizing, we have If f(x) > 0, its graph lies above the x-axis If f '(x) > 0, the graph of f(x) is increasing If f ''(x) > 0, the graph of f(x) is concave up 1) Given that f(x) = x 3 6x 2 + 5, determine the intervals where the graph: f ' x =3x 2 12 x => f '' x =6x 12 a. is concave upward. ( f ''(x) > 0 f is concave upward on ( 2, ) b. is concave downward. f is concave downward on ( -, 2) 2) Given f ''(x) = 5x 20, find where f ''(x) = 0 then find the intervals where f (x) is increasing and decreasing. f''(x) = 0 if x = 4 f '(x) is increasing if f ''(x) > 0, that is, on ( 4, ) f '(x) is decreasing if f ''(x) < 0, that is, on ( -, 4 )
3 Definition On an open interval (a, b) where f is differentiable: 1. If f ' (x) is increasing (f '' (x) > 0) then the graph of f(x) is concave up. 2. If f ' (x) is decreasing (f '' (x) < 0) then the graph of f(x) is concave down. Determine the intervals where the graph of f(x) = x 3 + 3x 2 4 is concave up and concave down. f ' x =3x 2 6x => f '' x =6x 6 => f '' x =0 if x = -1 f (x) is concave up if f ''(x) > 0, that is, on ( - 1, ) f (x) is concave down if f ''(x) < 0, that is, on ( -, - 1 ) Definition A point x=c in the domain of f is called an inflection value if f''' changes sign around it Note: The natural candidate to be an inflection point is one where f '' is zero or does not exist. Second Derivative Test (Determines inflection values and concavity) 1. Find the values x = c where f (c) = 0 or f (c) does not exist 2. Place these values on a number line and use the second derivative to generate a sign chart. 3. The point (c, f(c)) is an inflection point if f (x) changes sign at x = c and if x = c is in the domain of f(x).
4 1) Find the inflection points of the function f x = x 2 x 1 x 2 f x = x 2 x 1 x 2 = x 2 x 2 x 2 =x 4 x 3 2x 2 => f ' x =4x 3 3x 2 4x => f '' x =12x 2 6x 4 f ''(x) = 0 => 12x 2 6x 4=0 => 12x 2 6x 4=2 6x 2 3x 2 =0 x= 3± = 3± Thus the possible inflection points are : x = -3 - [ 57 / 12] and x = -3 + [ 57 / 12] < > f '' f is concave up on ( -, -3 - [ 57 / 12 ] ) ( -3 - [ 57 / 12], -3 + [ 57 / 12 ] ) f is concave down ( -3 + [ 57 / 12 ], ) 2) The Continental Company has determined that its cost, in hundreds of dollars, for producing x items of its best selling product is given by C(x) = x 3 6x x. (a) Determine C(5) and C (5) and interpret each. C 5 = =50 The cost of producing 5 items is $5000 C ' x =3x 2 12x 15 => C ' 5 = =30 The cost of producing the 6 st item is approximately $3000.
5 (b) Determine the intervals where C (x) is increasing and where it is decreasing. Determine the relative minimum for the marginal cost function. C'(x) is increasing if C''(x) > 0 and it is decreasing if C''(x) < 0 but C '' x =6x 12 therefore C'(x) is decreasing ( C'' < 0 ) on ( -, 2 ) and increasing ( C'' > 0 ) on ( 2, ). Thus, the marginal cost function C' has a minimum value at x= 2 (c) Determine the inflection point for the graph of C. From the above analysis we have that C has an inflection point at x= 2
6 Point of Diminishing Returns If a company expects increased dollars spent on advertising to increase sales. A the beginning the sales will increase at an increasing rate and then increase at a decreasing rate. The value of x where the rate of change of sales changes from increasing to decreasing is called the point of diminishing returns. A company estimates that it will sell N(x) units of a product after spending $x thousand dollars on advertising, as given by N x = 0.25x 4 13x 3 180x 2 10, x 24 When is the rate of change of sales increasing and when its decreasing. What is the point of diminishing returns and the maximum rate of change of sales? If N x = 0.25x 4 13x 3 180x 2 10,000, then N ' x = x 3 39x 2 360x and N '' x = 3x 2 78x 360= 3 x 2 26x 120 = 3 x 6 x < > f '' + 0 Thus, N' is increasing on ( 15, 20 ) and decreasing on ( 20, 24 ). Therefore the point of diminishing returns is x= 20 and the maximum rate of change is N'(2) = 400.
7 C) Curve Sketching To sketch the graph of a given function f(x) is good to keep in mind the following strategy. Analyze f(x) Find the domain of the function. Find the intercepts. Find the asymptotes. Analyze f (x) Find the zeros of f (x) and points where f '(x) does not exist Construct a sign chart for f (x), and determine where the function is increasing/decreasing Find the local extrema Analyze f (x). Find the zeros of f (x) and points where f ''(x) does not exist. Using a sign chart, determine where the graph of the function is concave up or down Find any points of inflection. Sketch the graph of the function. Sketch the graph of f(x) = x 4 + 2x 3. f x =x 4 2x 3 =x 3 x 2 => x-int x = - 2, x = 0 ; y-int y=0 ; asymptotes. f ' x =4x 3 6x 2 =2x 2 2x 3 => x = 0 and x = - 3/2 critical points f '' x =12x 2 12x=12x x 1 => x = 0 and x = - 1 possible inflection points
8 -2-3/ < > f ' rel min /2 0 1 < > f '' inf point inf point f is increasing up on ( - 3/2, 0 ) ( 0, 1 ) f is decreasing down ( -, - 3/2 ) f is concave up on ( -, -1) ( 0, ) f is concave down ( - 1, 0 )
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