Analyzing functions #3 - Questions and Answers

Question 1 Find the difference quotient of f, that is, find (f(x + h) – f(x))/h, h≠ 0, for each function. Be sure to simplify. f(x) = 1/(x+3) Answer f(x) = 1/(x+3) (f(x + h) - f(x))/h = (1/(x+h+3) - 1/(x+3))/h = ((x+3) - (x+h+3))/((x+h+3)(x+3)h) = (x+3-x-h-3)/((x+h+3)(x+3)h) = -h/((x+h+3)(x+3)h) = -1/((x+h+3)(x+3)) [Second attempt at calculating the difference quotient] f(x) = 1/(x+3) (f(x+h)-f(x))/h = (1/(x+h+3) - 1/(x+3))/h = (x+3-(x+h+3))/((x+h+3)(x+3)h) = (x+3-x-3-h)/((x+h+3)(x+3)h) = -h/((x+h+3)(x+3)h) = -1/((x+h+3)(x+3)) Question 2 What period do the sine, cosine, cosecant, and secant functions have? What period do the tangent and cotangent functions have? Answer The sine, cosine, cosecant, and secant functions repeat every 2π units, and the tangent and cotangent functions repeat every π units. Question 3 Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. f(x,y) = x³- 6xy + 8y³ Answer With the given function, determine the critical points and determine if the critical points are local minimum, maximum or saddle point. f(x,y) = x³- 6xy + 8y³ Identify first derivatives in terms of fₓ(x,y) and f_y(x,y). fₓ(x,y) = 3x² - 6y f_y(x,y) = -6x + 24y² Find the critical points of x by solving for y in fₓ and substituting result into f_y. fₓ(x,y) = 3x² - 6y = 0 => 3x² = 6y => x²/2 = y f_y(x,y) = -6x + 24(x²/2)² => -6x + 6x⁴ = 0 => 6x(x³ - 1) = 0 => x = 0 and x = 1 Find the critical values of y associated with the critical values found for x by substituting into y. f(0,y) = 3(0)² - 6y = 0 => y = 0. f(1,y) = 3(1)² - 6y = 0 => 3 = 6y => y = 1/2 Using fₓ(x,y) and f_y(x,y) find the second derivatives in terms of fₓₓ(x,y), f_yy(x,y) and f_xy(x,y) fₓₓ(x,y) = 6x f_yy(x,y) = 48y f_xy(x,y) = -6 Use the second derivative test provided the rules defined. D=D(x,y) = fₓₓ(x,y)f_yy(x,y) - (f_xy(x,y))² D>0 and fₓₓ(x,y) >0, then f(x,y) is local minimum D>0 and fₓₓ(x,y) <0, then f(x,y) is local maximum D<0 then f(x,y) is a saddle point The critical points identified as (0,0) and (1, 1/2) are to be used to solve [Image of work showing calculation of D for the function at the critical points] D=D(0,0) = (60)(480) - (-6)² = -36; D=D(1,1/2) = (61)(481/2) - (-6)² = 108 Compare D values of fₓₓ(x,y) to identify critical points as local maximum, minimum or saddle points D = D(0,0) = -36 => D<0, Saddle Point D = D(1,1/2) = 108, fₓₓ(1,1/2) > 0 D(1, 1/2) is local minimum Question 4 Let g(x) = ∫(from 0 to x) f(t)dt where f is the function whose graph is shown in the figure. (a) Estimate g(0), g(2), g(4), g(6), and g(8). (b) Find the largest open interval on which g is increasing. Find the largest open interval on which g is decreasing. (c) Identify any extrema of g. (d) Sketch a rough graph of g. Answer part (a) g(x) is the area between f and the x axis from 0 to x, with the area below the x axis being negative. We can estimate (get exact values actually) easily because the region is made up of triangles and rectangles. To find g(6) we add the triangle from (4,5) to the rectangle from (5,6) to g(4) which we calculated already. We can do the same thing for g(8) g(0) = 0, g(2) = 1/2 * (2)(-4) = -4 g(4) = 2 * (-4) = -8 g(6) = 1/2 (1)(4) + (1)(4) + g(4) = 2 + 4 -8= -2 g(8) = 1/2 * (1) * (4) + 1/2 * (1) * (4) + g(6) = 2 + 2 - 2 = 4 part (b) g is increasing when g'(which is f) is positive. Increasing on interval (4,8) Decreasing on (0,4) part (c) f changes from negative to positive at t=4 (corresponding to x=4) so there is a Relative minimum at that point in g. Relative minimum at x = 4 [Image of work showing the area calculations for g(x), the analysis of increasing and decreasing intervals, and the locations of extrema] [Image of a graph showing a rough sketch of g(x)] part (d) Graph of g(x). g is composed of piecewise linear functions (boundaries marked on graph) which you can calculate the exact values from using the graph of f and the various parts values above (the various parts values above For example, in interval [2,4] f(t) = 2t - 8 g(x) = x² - 8x + C g(4) = -8 = 4² - 8(4) + C C = -8 - 16 + 32 = 8 g(x) = x² - 8x + 8 You can do likewise for the other intervals. Result: (a) 0,-4,-8,-2,4; (b) Increasing on interval (4,8), Decreasing ON (0,4); (c) Relative Minimum at x=4; (d) See solution for graph Question 5 Explain the difference between an absolute minimum and a local minimum. There is no difference. A function f has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain. A function f has an absolute minimum at x = c if f(c) is the smallest function value when x is near c, whereas f has a local minimum at c if f(c) is the smallest function value on the entire domain of f. A function f has an absolute minimum at x = c if f(c) is the largest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the largest function value when x is near c. A function f has an absolute minimum at x = c if f(c) is the largest function value when x is near c, whereas f has a local minimum at c if f(c) is the largest function value on the entire domain of f. Answer Option 2 is correct: A function f has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain of f, whereas f has a local minimum at c if f(c) is the smallest function value when x is near c.