By Jon Rogawski
Read or Download Student Solutions Manual for Calculus Late Transcendentals Single Variable (Second Edition) PDF
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Additional info for Student Solutions Manual for Calculus Late Transcendentals Single Variable (Second Edition)
Sample text
Continue the graph of f (x) to the interval [−4, 4] as an even function. Sketch the graphs of y = 12 f (x) and y = f 12 x . solution To continue the graph of f (x) to the interval [−4, 4] as an even function, reflect the graph of f (x) across the y-axis (see the graph below). y 3 2 1 −4 −3 −2 −1 x 1 2 3 4 In Exercises 11–14, theof domain and of the function. Continue thefind graph f (x) to therange interval [−4, 4] as an odd function. √ 11. f (x) = x + 1 √ solution The domain of the function f (x) = x + 1 is {x : x ≥ −1} and the range is {y : y ≥ 0}.
05 m/s. 6 m/s. √ 3. Let v = 20 T as in Example 2. Estimate the instantaneous rate of change of v with respect to T when T = 300 K. 9t 2 m in t seconds. Estimate the solution instantaneous velocity at t = 3. 57735 m/(s · K). 31 June 7, 2011 LTSV SSM Second Pass 32 CHAPTER 2 LIMITS In Exercises 5 and 6,y/a stone tossed vertically the yair=from level withinstantaneous an initial velocity of change 15 m/s. of Its yheight Compute x foristhe interval [2, 5],into where 4x −ground 9. 9t m. 5. 5] and indicate the corresponding secant line on a sketch of the graph of h(t).
Find a formula for the average rate of change of f (x) = x 3 over [2, x] and use it to estimate the instantaneous rate of change at x = 2. solution The average rate of change is x3 − 8 f (x) − f (2) = . x−2 x−2 Applying the difference of cubes formula to the numerator, we find that the average rate of change is (x 2 + 2x + 4)(x − 2) = x 2 + 2x + 4 x−2 for x = 2. The closer x gets to 2, the closer the average rate of change gets to 22 + 2(2) + 4 = 12. √ Let T = 32 L as in Exercise 21. The numbers in the second column of Table 4 are increasing, and those in the last column are decreasing.