Math 181 - Quiz 1
Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) a) lim x → 2− f(x) b) lim x → 2+ f(x) c) lim x → 2 f(x) d) f(2) e) lim x → 4 f(x) f) f(4)
(a) As x approaches 2 from the left, the values of f(x) approach 3, so lim x → 2− f(x) = 3. (b) As x approaches 2 from the right, the values of f(x) approach 1, so lim x → 2+ f(x) = 1. (c) lim x → 2 f(x) does not exist since the left-hand limit does not equal the right-hand limit. (d) When x = 2, y = 3, so f(2) = 3. (e) As x approaches 4, the values of f(x) approach 4, so lim x → 4 f(x) = 4. (f) There is no value of f(x) when x = 4, so f(4) does not exist.
The table shows the position of a cyclist. see pic for graph (a) Find the average velocity for each time period. (b) Estimate the instantaneous velocity when t = 3.
(a) Find the average velocity for each time period. (i) On the interval [1, 3], vavg = (s(3) − s(1)) / (3 − 1) = 10.3 − 1.5 / 2 = 8.8 2 = 4.4 m/s. (ii) On the interval [2, 3], vavg = (s(3) − s(2)) / (3 − 2) = 10.3 − 4.8 /1 = 5.5 1 = 5.5 m/s. (iii) On the interval [3, 5], vavg = (s(5) − s(3)) / (5 − 3) = 26.8 − 10.3 / 2 = 16.5 2 = 8.25 m/s. (iv) On the interval [3, 4], vavg = (s(4) − s(3)) / (4 − 3) = 17.9 − 10.3 / 1 = 7.6 / 1 = 7.6 m/s. (b) The instantaneous velocity when t = 3 is about 6.55 m/s.
A graphing calculator is recommended. A bacteria culture starts with 700 bacteria and doubles in size every half hour. (a) How many bacteria are there after 4 hours? (b) How many bacteria are there after t hours? (c) How many bacteria are there after 40 minutes? (Round your answer to the nearest whole number.) (d) Graph the population function. (e) Estimate the time for the population to reach 20,000. (Round your answer to one decimal place.)
(a) Four hours represents 8 doubling periods (one doubling period is 30 minutes). 700 · 2 8 = 179, 200 (b) In t hours, there will be 2t doubling periods. The initial population is 700, so the population y at time t is y = 700 · 2 2t = 700 · 4 t . (c) t = 40 60 = 2 3 ⇒ y = 700 · 2 2(2/3) = 1764. (d) see pic (e) We graph y1 = 700 · 2 2t and y2 = 20, 000. The two curves intersect at t = 2.4, so the population reaches 20, 000 in about 2.4 hours.
The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $390 to drive 300 mi and in June it cost her $558 to drive 860 mi. (a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model. (b) Use part (a) to predict the cost of driving 1800 miles per month. Draw the graph of the linear function. (c) What does the slope represent? (d) What does the C-intercept represent? (e) Why does a linear function give a suitable model in this situation?
(a) Using d in place of x and C in place of y, we find the slope to be C2 − C1 / d2 − d1 = 558 − 390 / 860 − 300 = 168 / 560 = 3 / 10 . So a linear equation is C−558 = 3 / 10 (d − 860) ⇔ C−558 = 3 / 10 d−258 ⇔ C = 0.3d + 300. (b) Letting d = 1800 we get C = 3 / 10 (1800)+300 = 840. The cost of driving 1800 miles is 840 dollars. see pic (c) The slope of the line represents the cost per mile, 30 cents/mile. (d) The y-intercept represents the fixed cost, 300 dollars. (e) A linear function gives a suitable model in this situation because you have fixed monthly costs such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for each additional mile driven is a constant.
A function is given by a verbal description. Determine whether it is one-to-one. The function f(t) is the height of a football t seconds after kickoff.
A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus, even if t1 does not equal t2, f(t1) may equal f(t2), so f is not 1-1.
Use the given graph of f to find a number δ such that if 0<abs x-3 < δ then abs f(x)-2 < 0.5
If |f(x) − 2| < 0.5, then −0.5 < f(x) − 2 < 0.5 1.5 < f(x) < 2.5. From the graph, we see that the last inequality is true if 2.6 < x < 3.8, so we can choose δ = min{3 − 2.6, 3.8 − 3} = min{0.4, 0.8} = 0.4 (or any smaller positive number). Note that x ≠ 3.
Find the limit. lim x→∞ (e−8x cos(x))
Since −1 ≤ cos(x) ≤ 1 and e −8x > 0, we have −e −8x ≤ e −8x cos(x) ≤ e −8x . We know that limx→∞ (−e −8x ) = 0 and limx→∞ (e −8x ) = 0, so by the Squeeze Theorem, limx→∞ (e −8x cos(x)) = 0.
If a rock is thrown upward on the planet Mars with a velocity of 17 m/s, its height (in meters) after t seconds is given by H = 17t − 1.86t2. (a) Find the velocity of the rock after two seconds. (b) Find the velocity of the rock when t = a. (c) When will the rock hit the surface? (Round your answer to one decimal place.) (d) With what velocity will the rock hit the surface?
a) Let H(t) = 17t − 1.86t 2 . v(2) = lim h→0 H(2 + h) − H(2) h = lim h→0 [17(2 + h) − 1.86(2 + h) 2 ] − (34 − 7.44) h = lim h→0 34 + 17h − 1.86(4 + 4h + h 2 ) − 34 + 7.44 h = lim h→0 34 + 17h − 7.44 − 7.44h − 1.86h 2 − 34 + 7.44 h = lim h→0 9.56h − 1.86h 2 h = lim h→0 (9.56 − 1.86h) = 9.56 The velocity of the rock after one second is 9.56 m/ s. (b) Let H(t) = 17t − 1.86t 2 . v(a) = lim h→0 H(a + h) − H(a) h = lim h→0 [17(a + h) − 1.86(a + h) 2 ] − (17a − 1.86a 2 ) h = lim h→0 17a + 17h − 1.86(a 2 + 2ah + h 2 ) − 17a + 1.86a 2 h = lim h→0 17a + 17h − 1.86a 2 − 3.72ah − 1.86h 2 − 17a + 1.86a 2 h = lim h→0 17h − 3.72ah − 1.86h 2 h = lim h→0 h(17 − 3.72a − 1.86h) h = lim h→0 (17 − 3.72a − 1.86h) = 17 − 3.72a The velocity of the rock when t = a is (17 − 3.72a) m/ s. (c) The rock will hit the surface when H = 0 ⇔ 17t − 1.86t 2 = 0 ⇔ t(17 − 1.86t) = 0 ⇔ t = 0 or 1.86t = 17. The rock hits the surface when t = 17/1.86 ≈ 9.1 s. (d) The velocity of the rock when it hits the surface is v (17 /1.86) = 17 − 3.72 (17/ 1.86) = 17 − 34 = −17 m/s
Consider the following. f(x) = e^x if x < 0 = x^2 if x ≥ 0 , a = 0 a) Find the left-hand and right-hand limits at the given value of a. lim x→0− f(x) = lim x→0+ f(x) = b) Explain why the function is discontinuous at the given number a. Since these limits are________lim x→0 f(x)_________and f is therefore discontinuous at 0. c) Sketch the graph of the function.
a) The left-hand limit of f at a = 0 is lim x → 0− f(x) =lim x → 0− ex = 1. The right-hand limit of f at a = 0 is lim x → 0+ f(x) = lim x → 0+ x2 = 0. b) Since these limits are not equal, lim x → 0 f(x) does not exist and f is discontinuous at 0. c) see pic
The graph of f is given. State the numbers at which f is not differentiable. x= (smaller) x= (bigger)
f is not differentiable at x = −1, because there is a discontinuity there, and at x = 2, because the graph has a corner there.
From the graph of g, state the intervals on which g is continuous.
g is continuous on [−4, −2),(−2, 2), [2, 4),(4, 6),(6, 8).
Evaluate the limit using appropriate Limit Laws lim x-->-2 (2x^2 + 2) / (x^2 + 2x -1)
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The displacement (in meters) of a particle moving in a straight line is given by s=t^2-9t+15, where t is measured in seconds. a) find the average velocity over each time interval b) find the instantaneous velocity when t=4
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what are the limit laws
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The graphs of four derivatives are given below. Match the graph of each function in (a)-(d) with the graph of its derivative in I-IV.
see pic for answer (a)′ = I, since from left to right, the slopes of the tangents to the graph start out negative, become 0, then positive, then 0 , then negative again. The actual function values in the graph follow the same pattern. (b)′ = III, since from left to right, the slopes of the tangents to the graph start out at a fixed positive quantity, then suddenly become negative, then positive again. The discontinuities in the graph indicate sudden changes in the slopes of the tangents. (c)′ = IV, since the slopes of the tangents to the graph are negative for x < 0 and positive for x > 0, as are the function values of the graph. (d)′ = II, since from left to right, the slopes of the tangents to the graph are positive, then 0, then negative, then 0, then positive, then 0, then negative again, and the function values in the graph follow the same pattern.