Applications of Differentiation PDF

CSCI 2305 Mathematics for Computing II  Maximum and minimum values  Increasing and Decreasing functions  Concavity and the Second Derivative test A number c in the domain of a function f is called a critical number of f if f’(c) = 0 or f’(c) is undefined....

CSCI 2305 Mathematics for Computing II  Maximum and minimum values  Increasing and Decreasing functions  Concavity and the Second Derivative test A number c in the domain of a function f is called a critical number of f if f’(c) = 0 or f’(c) is undefined. Slide 3 Find the critical numbers and local extrema of SOLUTION Critical numbers: The critical numbers x = –1 and x = 2 correspond to local maxima and local minima. Slide 4  Find the critical numbers and local extrema of f (x) = (3x + 1)2/3. SOLUTION Of course, f’ (x) ≠ 0 for all x, but f (x) is undefined at x = −1/3 , which is in the domain of f. Thus, x = −1/3 is the only critical number of f. x = –1/3 corresponds to the location of a local minimum (also the absolute minimum). Slide 5 Find the critical numbers and local extrema of f (x) = x3. SOLUTION f has a horizontal tangent line at x = 0, but does not have a local extremum there. Slide 6 Find the critical numbers and local extrema of f (x) = x1/3. SOLUTION Slide 7 SOLUTION Note that the domain of f consists of all real numbers other than x = −2. f’(x) = 0 for x = 0,−4 and f’(x) is undefined for x = −2. However, −2 is not in the domain of f and consequently, the only critical numbers are x = 0 and x = −4. Slide 8 Local extremum occurs either at 1. at a point where the tangent line is horizontal) [i.e., f’(x) = 0], 2. at a point where the tangent line is vertical [where f’(x) is undefined] 3. or at a corner [again, where f’(x) is undefined]. Slide 9 Locate any local extrema for f (x) = 9 − x2 and describe the behavior of the derivative at the local extremum. There is a local maximum at x = 0. Slide 10 Locate any local extrema for f (x) = x2 + 5x − 1 and describe the behavior of the derivative at the local extremum. There is a local minimum at x =. Slide 11 Locate any local extrema for f (x) = −x2 + 4x + 2 and describe the behavior of the derivative at the local extremum. There is a local minimum at x =. Slide 12 Locate any local extrema for f (x) = x4 − 2x2 + 1 and describe the behavior of the derivative at the local extremum. There is a local minimum at x =. There is a local maximum at x =. Slide 13 Locate any local extrema for f (x) = |x| and describe the behavior of the derivative at the local extremum. There is a local minimum at x = 0. The graph has a corner at x = 0 and hence, f’ (0) is undefined. Slide 14 For a function f defined on a set S of real numbers and a number c ∈ S, (i) f (c) is the absolute maximum of f on S if f (c) ≥ f (x) for all x ∈ S and (ii) f (c) is the absolute minimum of f on S if f (c) ≤ f (x) for all x ∈ S. Slide 15 (a) Locate any absolute extrema of f (x) = x2 − 9 on the interval (−∞,∞). (b) Locate any absolute extrema of f (x) = x2 − 9 on the interval (−3, 3). (c) Locate any absolute extrema of f (x) = x2 − 9 on the interval [−3, 3]. Slide 16 A Function with No Absolute Maximum or Minimum. Locate any absolute extrema of f (x) = 1/x, on [−3, 0) ∪ (0, 3]. No Absolute Maximum or Minimum f clearly fails to have either an absolute maximum or an absolute minimum on [−3, 0) ∪ (0, 3]. Slide 17 (Extreme Value Theorem) A continuous function f defined on a closed, bounded interval [a, b] attains both an absolute maximum and an absolute minimum on that interval. Find the absolute extrema of f (x) = 1/x on the interval [1, 3]. On the interval [1, 3], f is continuous. Judging from the graph in the figure, it appears that f (x) reaches its maximum value of 1 at x = 1 and its minimum value of 1/3 at x = 3. Slide 18 Theorem 3.3 gives us a simple procedure for finding the absolute extrema of a continuous function on a closed, bounded interval: 1. Find all critical numbers in the interval and compute function values at these points. 2. Compute function values at the endpoints. The largest function value is the absolute maximum and the smallest function value is the absolute minimum. Slide 19 Find the absolute extrema of f (x) = 2x3 − 3x2 − 12x + 5 on the interval [−2, 4]. SOLUTION The maximum appears to be at the endpoint x = 4. The minimum appears to be at a local minimum near x = 2. The critical numbers are x = −1 and x = 2. Both of these are in the interval [−2, 4]. Compare the values at the endpoints: and the values at the critical numbers: Thus, f (4) = 37 is the absolute maximum and f (2) = −15 is the absolute minimum. Slide 20 A number c in the domain of a function f is called a critical number of f if f’(c) = 0 or f’(c) is undefined. (Fermat’s Theorem) Suppose that f (c) is a local extremum (local maximum or local minimum). Then c must be a critical number of f. Suppose that f is continuous on the closed interval [a, b]. Then, each absolute extremum of f must occur at an endpoint (a or b) or at a critical number. Slide 21  A function f is increasing on an interval I if for every x1, x2 ∈ I with x1 < x2, f (x1) < f (x2) [i.e., f (x) gets larger as x gets larger]. A function f is decreasing on the interval I if for every x1, x2 ∈ I with x1 < x2, f (x1) > f (x2) [i.e., f (x) gets smaller as x gets larger]. Local extrema occur at points where a function changes from increasing to decreasing or vise versa. Slide 22 Suppose that f is continuous on the interval [a, b] and c ∈ (a, b) is a critical number. (i) If f‘(x) > 0 for all x ∈ (a, c) and f’(x) < 0 for all x ∈ (c, b) (i.e., f changes from increasing to decreasing at c), then f(c) is a local maximum. (ii) If f’(x) < 0 for all x ∈ (a, c) and f’(x) > 0 for all x ∈ (c, b) (i.e., f changes from decreasing to increasing at c), then f(c) is a local minimum. (iii) If f’(x) has the same sign on (a, c) and (c, b), then f‘(c) is not a local extremum. Slide 23  Find where the function f (x) = 2x3 + 9x2 − 24x − 10 is increasing and where it is decreasing. Solution:  Find where the function f (x) = 3x4 – 4x3 – 12x2 + 5 is increasing and where it is decreasing. Solution: f ′(x) = 12x3 – 12x2 – 24x = 12x(x – 2)(x + 1) Solution: Find the local extrema of f (x) = x4 + 4x3 − 5x2 − 31x + 29 and draw a graph. SOLUTION We can find approximations to the three zeros/roots of f’ : a ≈ −2.96008, b ≈ −1.63816 and c ≈ 1.59824. There is a local minimum at a ≈ −2.96008, a local maximum at b ≈ −1.63816 and a local (and absolute) minimum at c ≈ 1.59824. Slide 27 Slide 28 Discuss the curve y = x4 – 4x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. Solution: If f (x) = x4 – 4x3, then f ′(x) = 4x3 – 12x2 = 4x2(x – 3) \ f ″ (x) = 12x2 – 24x = 12x(x – 2) Slide 29 Inflection points, (-3/2 , f (-3/2 ) Slide 30 Slide 31  Sketch the graph of the function f (x) = x2/3(6 – x)1/3  Solution: Calculation of the first two derivatives gives Since f ′(x) = 0 when x = 4 and f ′ (x) does not exist when x = 0 or x = 6, the critical numbers are 0, 4 and 6.