SOLUTION MANUALS FOR CALCULUS EARLY TRANSCENDENTALS 9th EDITION BY JAMES STEWART, DANIEL K. CLEGG & SALEEM WATSON 2023/2024 / GRADED A+.

SOLUTION MANUALS FOR CALCULUS EARLY TRANSCENDENTALS 9th EDITION BY JAMES STEWART, DANIEL K. CLEGG & SALEEM WATSON 2023/2024 / GRADED A+. 1 FUNCTIONS AND MODELS 1.1 Four Ways to Represent a Function 1. The functions () =  + √ 2 − and () =  + √ 2 − give exactly the same output values for every input value, so and are equal. 2. () =2 −   − 1 = ( − 1)  − 1 = for  − 1 6= 0, so and [where () = ] are not equal because  (1)is undefined and (1) = 1. 3. (a) The point (−2 2)lies on the graph of, so (−2) = 2. Similarly,  (0) = −2,  (2) = 1, and  (3)  2. 5 (b) Only the point (−4 3)on the graph has a value of 3, so the only value offor which () = 3is −4 . (c) The function outputs (are never greater than 3, so ) () ≤ 3for the entire domain of the function. Thus,() ≤ 3for −4 ≤  ≤ 4 (or, equivalently, on the interval[−4 4]). (d) The domain consists of all values on the graph of : { | −4 ≤  ≤ 4} = [−4 4]. The range of consists of all the values on the graph of : { | −2 ≤  ≤ 3} = [−2 3]. (e) For any 1   2 in the interval [0 2] , we have (1)  (2). [The graph rises from (0 −2)to (2 1).] Thus, (is) increasing on [0 2] . 4. (a) From the graph, we have (−4) = −2 and (3) = 4. (b) Since (−3) = −1 and (−3) = 2, or by observing that the graph ofis above the graph ofat  = −3, (−3)is larger than (−3). (c) The graphs of and intersect at  = −2 and  = 2, so  () = (at these two values of ) . (d) The graph of lies below or on the graph offor −4 ≤  ≤ −2 and for 2 ≤  ≤ 3. Thus, the intervals on which  () ≤ (are) [−4 −2]and [2 3] . (e)  () = −1is equivalent to = −1, and the points on the graph ofwith values of −1 are (−3 −1)and (4 −1), so the solution of the equation() = −1is  = −3 or  = 4. (f) For any 1  2 in the interval [−4 0], we have (1)  ( 2 ). Thus, (is decreasing on ) [−4 0]. (g) The domain of is { | −4 ≤  ≤ 4} = [−4 4]. The range of is { | −2 ≤  ≤ 3} = [−2 3]. (h) The domain of is { | −4 ≤  ≤ 3} = [−4 3]. Estimating the lowest point of the graph ofas having coordinates (0 0, the range of 5) is approximately { | 05 ≤  ≤ 4} = [05. 4] 5. From Figure 1 in the text, the lowest point occurs at about(   ) = (12 −. The highest point occurs at about 85) (17 115) . Thus, the range of the vertical ground acceleration is−85 ≤  ≤ 115. Written in interval notation, the range is[−85 115