University:
University of Michigan
Course:
MATH 115 | Calculus I
Academic year:

2021
Views:

500

Pages:

1
Author:

Draven Cohen

Math 115 / Exam 1 (February 23, 2021 ) page 14 10. [14 points] The graph of the function f (x) is shown below. y (Reduced scale for solutions) 4 y = f (x) 3 2 1 −4 −3 −2 −1 −1 1 2 3 x −2 For a.–b., give your answers as a list of one or more of the given numbers, or write none a. [2 points] For which of the values c = −3, −2, −1, 0, 1 is f (x) continuous at x = c? Solution: Among these options, f (x) is only continuous when x = −3 and when x = −1. b. [2 points] For which of the values c = −3, −2, −1, 0, 1 is lim f (x) = f (c) ? x→c− Solution: This condition is certainly satisfied at the points where f (x) is continuous, so we already know that x = −3 and x = −1 should be in our answer. Checking the other points: lim f (x) = −1 and f (−2) = −1. lim f (x) = 4 but f (0) = −2. lim f (x) = −2 but f (1) = 3. x→−2− x→0− x→1− So, our final answer is that the desired condition is satisfied at x = −3, − 2, and − 1 . For c.–g., use the graph of the function f (x) to evaluate each of the expressions below. If a limit diverges to ∞ or −∞ or if the limit does not exist for any other reason, write “dne.” If there is not enough information to evaluate the expression, write “Not enough information.” c. [2 points] lim f (x) x→0 Solution: Observe that limx→0− f (x) = 4 and limx→0+ f (x) = −2. Since the left- and righthand limits are different, we see that limx→0 f (x) does not exist, so we write “DNE.” d. [2 points] lim f (x) x→1 Solution: On either side of x = 1, the function f (x) is constant and equal to −2. Therefore lim f (x) = -2 . x→1 e. [2 points] lim h→0 f (−1 + h) − f (−1) h Solution: We recognize this expression as f ′ (−1). Since the function f (x) is linear on the interval (−2, 0), we see that f ′ (−1) is the slope of this line. We calculate the slope to be (4 − 1)/(0 − (−2)) = 3/2. f. [2 points] lim 4f (x − 5) − 1 x→3+ Solution: As x approaches 3 from the right, x − 5 approaches 2 from the right. This means that f (x − 5) approaches 1 from above, and therefore 4f (x − 5) − 1 approaches 4(1) − 1 = 3 . g. [2 points] lim f (f (x)) x→−3 Solution: As x approaches −3 from either side, the value y = f (x) approaches 0 from negative values only. Therefore lim f (f (x)) = lim f (y) = 4 . x→−3 y→0−