Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = 1 x+5 a = −5 f(−5) is undefined. limx→−5 f(x) does not exist. limx→−5+f(x) and limx→−5−f(x) are finite, but are not equal. f(−5) is defined and limx→−5 f(x) is finite, but they are not equal. none of the above Sketch the graph of the function.
For the function h whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) (a) limx→−3−h(x) (b) limx→−3+h(x) (c) limx→−3 h(x) (d) h(−3) (e) limx→0−h(x) (f) limx→0+h(x) (g) limx→0 h(x) (h) h(0) (i) limx→2 h(x) (j) h(2) (k) limx→5+h(x) (l) limx→5−h(x)
The signum (or sign) function, denoted by sgn, is defined by sgnx = {−1 if x < 0 0 if x = 0 1 if x > 0 (a) Sketch the graph of this function. (b) Find each of the following limits. (If an answer does not exist, enter DNE.) (i) limx→0+sgnx (ii) limx→0−sgnx (iii) limx→0 sgnx (iv) limx→0|sgnx|
Let g(x) = {x if x < 12 if x = 12−x2 if 1 < x ≤ 2 x−2 if x > 2. (a) Evaluate each of the following, if it exists. (If an answer does not exist, enter DNE.) (i) limx→1−g(x) (ii) limx→1+g(x) (iii) g(1) (iv) limx→2−g(x) (v) limx→2+g(x) (vi) limx→2 g(x) (b) Sketch the graph of g.
The toll T charged for driving on a certain stretch of a toll road is $5 except during rush hours (between 7 AM and 10 AM and between 4 PM and 7 PM) when the toll is $7. (a) Sketch a graph of T as a function of the time t, measured in hours past midnight. (b) Locate the discontinuities of T. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) t = Classify the discontinuities as removable, jump, or infinite. removable jump infinite none −T is continuous Discuss the significance of the discontinuities of T to someone who uses the road. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 0 and t = 7, between t = 10 and t = 16, and between t = 19 and t = 24 if feasible. The function is continuous, so there is no significance. Because of the sudden jumps in the toll, drivers may want to avoid the higher rates between t = 7 and t = 10 and between t = 16 and t = 19 if feasible. Because of the steady increases and decreases in the toll, drivers may want to avoid the highest rates at t = 7 and t = 24 if feasible.
Find the domain of f and f−1 and its domain. f(x) = ln(ex−4). (a) Find the domain of f. (Enter your answer using interval notation.) (b) Find f−1. f−1(x) = Find its domain. (Enter your answer using interval notation.)
If a bacteria population starts with 100 bacteria and doubles every three hours, then the number of bacteria after t hours is n = f(t) = 100⋅2 t/3 (a) Find the inverse of this function. f−1(n) = Explain its meaning in the context of this problem. This function tells us how long it will take to obtain n bacteria (given the number t). This function tells us how long it will take to obtain n bacteria (given the number n). (b) When (in hr) will the population reach 10, 000? (Round your answer to one decimal place.) hr
If a bacteria population starts with 80 bacteria and doubles every three hours, then the number of bacteria after t hours is n = f(t) = 80⋅2 t/3 (a) Find the inverse of this function. f−1(n) = Explain its meaning. This function tells us how long it will take to obtain n bacteria (given the number t). This function tells us how long it will take to obtain n bacteria (given the number n). (b) When will the population reach 70, 000? (Round your answer to one decimal place.) hr
The half-life of cesium-137 is 30 years. Suppose we have a 130−mg sample. (a) Find the mass that remains after t years. y(t) = mg (b) How much of the sample remains after 100 years? (Round your answer to two decimal places.) mg (c) After how long will only 1 mg remain? (Round your answer to one decimal place.) t = yr
For the function f whose graph is shown, state the following. (If an answer does not exist, enter DNE.) (a) limx→−7 f(x) −∞ (b) limx→−3 f(x) ∞ (c) limx→0 f(x) ∞ (d) limx→6−f(x) −∞ (e) limx→6+f(x) ∞ (f) The equations of the vertical asymptotes. x = (smallest value) x = x = x = (largest value)
Explain what is meant by the equation limx→3 f(x) = 5. If |x1−3| < |x2−3|, then |f(x1)−5| ≤ |f(x2)−5|. The values of f(x) can be made as close to 5 as we like by taking x sufficiently close to 3. The values of f(x) can be made as close to 3 as we like by taking x sufficiently close to 5. If |x1−3| < |x2−3|, then |f(x1)−5| < |f(x2)−5|. f(x) = 5 for all values of x. Is it possible for this statement to be true and yet f(3) = 8? Explain. Yes, the graph could have a hole at (3, 5) and be defined such that f(3) = 8. Yes, the graph could have a vertical asymptote at x = 3 and be defined such that f(3) = 8. No, if f(3) = 8, then limx→3 f(x) = 8. No, if limx→3 f(x) = 5, then f(3) = 5.
Under ideal conditions a certain bacteria population is known to double every three hours. Suppose there are initially 140 bacteria. (a) What is the size of the population after 12 hours? (b) What is the size of the population after t hours? (c) What is the size of the population after 19 hours? (Round your answer to the nearest whole number.) (d) Graph the population function and estimate the time for the population to reach 50, 000. (Round your answer to two decimal places.) hours
The point P(9, −3) lies on the curve y = 3/(8−x). (a) If Q is the point (x, 3/(8−x)), use your calculator to find the slope mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 8.9 mPQ = (ii) 8.99 mPQ = (iii) 8.999 mPQ = (iv) 8.9999 mPQ = (v) 9.1 mPQ = (vi) 9.01 mPQ = (vii) 9.001 mPQ = (viii) 9.0001 mPQ = (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(9, −3). m = (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(9, −3).
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {x2−4x x2−16 if x≠41 if x = 4 a = 4 limx→4+f(x) and limx→4−f(x) are finite, but are not equal. f(4) is defined and limx→4 f(x) is finite, but they are not equal. f(4) is undefined. limx→4 f(x) does not exist. none of the above Sketch the graph of the function.
Consider the following function. f(x) = ln(x)+7 (a) What are the domain and range of f? (Enter your answers using interval notation.) domain range (b) What is the x-intercept of the graph of f? x = (c) Sketch the graph of f.
Let g(x) = x2+x−12 |x−3| (a) Find the following limits. (If an answer does not exist, enter DNE.) (i) limx→3+g(x) (ii) limx→3−g(x) (b) Does limx→3 g(x) exist? Yes No (c) Sketch the graph of g.
Let f(x) = {1+x if x < −1 x2 if −1 ≤ x < 1 2−x if x ≥ 1 Sketch the graph of f. (i) Use the graph to determine the values of a for which limx→af(x) does not exist. (Enter your answers as a comma-separated list.) a =
Explain what it means to say that limx→7−f(x) = 4 and limx→7+f(x) = 1 As x approaches 7 from the right, f(x) approaches 4. As x approaches 7 from the left, f(x) approaches 1. As x approaches 7, f(x) approaches 4, but f(7) = 1. As x approaches 7, f(x) approaches 1, but f(7) = 4. As x approaches 7 from the left, f(x) approaches 4. As x approaches 7 from the right, f(x) approaches 1. In this situation is it possible that limx→7 f(x) exists? Explain. Yes, f(x) could have a hole at (7, 4) and be defined such that f(7) = 1. Yes, f(x) could have a hole at (7, 1) and be defined such that f(7) = 4. Yes, if f(x) has a vertical asymptote at x = 7, it can be defined such that limx→7−f(x) = 4, limx→7+f(x) = 1, and limx→7 f(x) exists. No, limx→7 f(x) cannot exist if limx→7−f(x)≠limx→7+f(x).
The point P(6, −3) lies on the curve y = 3/(5−x). (a) If Q is the point (x, 3 /(5−x)), use your calculator to find the slope mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 5.9 mPQ = (ii) 5.99 mPQ = (iii) 5.999 mPQ = (iv) 5.9999 mPQ = (v) 6.1 mPQ = (vi) 6.01 mPQ = (vii) 6.001 mPQ = (viii) 6.0001 mPQ = (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(6, −3). m = (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(6, −3).
The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. (a) Find the mass that remains after t years. y(t) = mg (b) How much of the sample remains after 150 years? (Round your answer to two decimal places.) y(150) = mg (c) After how long will only 1 mg remain? (Round your answer to one decimal place.) t = years
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {cos(x) if x < 0 0 if x = 0 1−x2 if x > 0 a = 0 limx→0+f(x) and limx→0−f(x) are finite, but are not equal. f(0) and limx→0 f(x) are finite, but are not equal. f(0) is undefined. limx→0 f(x) does not exist. none of the above Sketch the graph of the function.
The number N of bacteria in a culture after t days is modeled by N = 200[1 − 3 (t2+2)2]. Find the rate of change, in bacteria per day, of N with respect to t when the following values are true. (Round your answers to the nearest tenth.) (a) t = 0 bacteria per day (b) t = 1 bacteria per day (c) t = 2 bacteria per day (d) t = 3 bacteria per day (e) t = 4 bacteria per day (f) What can you conclude? After the first day the rate of change of the population is decreasing. The rate of change of the population is always increasing. The rate of change of the population is always decreasing. After the first day the rate of change of the population is increasing. The number of bacteria is only increasing during the first day.
The point P(8, −2) lies on the curve y = 2 7−x. (a) If Q is the point (x, 2 7−x), find the slope of the secant line PQ (correct to six decimal places) for the following values of x. (i) 7.9 mPQ = (ii) 7.99 mPQ = (iii) 7.999 mPQ = (iv) 7.9999 mPQ = (v) 8.1 mPQ = (vi) 8.01 mPQ = (vii) 8.001 mPQ = (viii) 8.0001 mPQ = (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(8, −2). m = (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(8, −2).
Explain what is meant by the equation limx→9 f(x) = 5 f(x) = 5 for all values of x. The values of f(x) can be made as close to 9 as we like by taking x sufficiently close to 5. If |x1−9| < |x2−9|, then |f(x1)−5| < |f(x2)−5|. If |x1−9| < |x2−9|, then |f(x1)−5| ≤ |f(x2)−5|. The values of f(x) can be made as close to 5 as we like by taking x sufficiently close to 9. Is it possible for this statement to be true and yet f(9) = 6? Explain. Yes, the graph could have a hole at (9, 5) and be defined such that f(9) = 6. Yes, the graph could have a vertical asymptote at x = 9 and be defined such that f(9) = 6. No, if f(9) = 6, then limx→9 f(x) = 6. No, if limx→9 f(x) = 5, then f(9) = 5.
f(x) = {x2 + 1 if x < 1 (x−2)2 if x ≥ 1 (a) Find the following limits. (If an answer does not exist, enter DNE.) limx→1−f(x) = limx→1+f(x) = (b) Does limx→1 f(x) exist? Yes No (c) Sketch the graph of f.
Explain what is meant by the equation limx→1 f(x) = 8 f(x) = 8 for all values of x. If |x1−1| < |x2−1|, then |f(x1)−8| < |f(x2)−8|. The values of f(x) can be made as close to 1 as we like by taking x sufficiently close to 8. The values of f(x) can be made as close to 8 as we like by taking x sufficiently close to 1. If |x1−1| < |x2−1|, then |f(x1)−8| ≤ |f(x2)−8|. Is it possible for this statement to be true and yet f(1) = 4? Explain. Yes, the graph could have a hole at (1, 8) and be defined such that f(1) = 4. Yes, the graph could have a vertical asymptote at x = 1 and be defined such that f(1) = 4. No, if f(1) = 4, then limx→1 f(x) = 4. No, if limx→1 f(x) = 8, then f(1) = 8.
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {x2−4x x2−16 if x ≠ 41 if x = 4 a = 4 f(4) is defined and limx→4 f(x) is finite, but they are not equal. f(4) is undefined. limx→4 f(x) does not exist. limx→4+f(x) and limx→4−f(x) are finite, but are not equal. none of the above Sketch the graph of the function.
When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by the following. Q(t) = Q0(1 − e−t/a) (The maximum charge capacity is Q0 and t is measured in seconds.) (a) Find the inverse of this function. t(Q) = Explain its meaning. This gives us the time t necessary to obtain a given charge Q. This gives us the time t with respect to the maximum charge capacity Q0. This gives us the charge Q obtained within a given time t. (b) How long does it take to recharge the capacitor to 77% of capacity if a = 5? (Round your answer to one decimal place.) sec
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {cos(x) if x < 0 0 if x = 0 1−x2 if x > 0 a = 0 limx→0+f(x) and limx→0−f(x) are finite, but are not equal. f(0) is undefined. limx→0 f(x) does not exist. f(0) and limx→0 f(x) are finite, but are not equal. none of the above Sketch the graph of the function.
Explain what is meant by the equation limx→7 f(x) = 4 If |x1−7| < |x2−7|, then |f(x1)−4| ≤ |f(x2)−4|. The values of f(x) can be made as close to 7 as we like by taking x sufficiently close to 4. f(x) = 4 for all values of x. If |x1−7| < |x2−7|, then |f(x1)−4| < |f(x2)−4|. The values of f(x) can be made as close to 4 as we like by taking x sufficiently close to 7. Is it possible for this statement to be true and yet f(7) = 8? Explain. Yes, the graph could have a hole at (7, 4) and be defined such that f(7) = 8. Yes, the graph could have a vertical asymptote at x = 7 and be defined such that f(7) = 8. No, if f(7) = 8, then limx→7 f(x) = 8. No, if limx→7 f(x) = 4, then f(7) = 4
Given that limx→1 f(x) = 4 limx→1 g(x) = −4 limx→1 h(x) = 0, find the limits, if they exist. (If an answer does not exist, enter DNE.) (a) limx→1 [f(x) + 5g(x)] (b) limx→1 [g(x)]3 (c) limx→1 f(x) (d) limx→1 2f(x) g(x) (e) limx→1 g(x) h(x) (f) limx→1 g(x)h(x) f(x)
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {1 x+3 if x ≠ −3 1 if x = −3 a = −3 f(−3) is defined and limx→−3 f(x) is finite, but they are not equal. f(−3) is undefined. limx→−3 f(x) is not finite. limx→ −3+f(x) and limx→−3 −f(x) are finite, but are not equal. none of the above Sketch the graph of the function.
Explain why the function is discontinuous at the given number a. (Select all that apply.) f(x) = {2x2−9 x−5 x−5 if x ≠ 5 10 if x = 5 a = 5 limx→5+ f(x) and limx→5− f(x) are finite, but are not equal. f(5) and limx→5 f(x) are finite, but are not equal. limx→5 f(x) does not exist. f(5) is undefined. none of the above Sketch the graph of the function.
Sketch the graph of the function. f(x) = {4+x if x < −2 x2 if −2 ≤ x < 2 6−x if x ≥ 2 Use the graph to determine the values of a for which limx→a f(x) does not exist. (Enter your answers as a comma-separated list.)
Video Example EXAMPLE 11 Show that the following limit is true. limx→0 x8sin(1 x) SOLUTION First note that we cannot use limx→0 x8 sin(1 x) = limx→0 x8⋅limx→0 sin(1 x) because the limit as x approaches 0 of sin(1 /x) does not exist (see this example). Instead we apply the Squeeze Theorem, and so we need to find a function f smaller than g(x) = x8 sin(1 /x) and a function h bigger than g such that both f(x) and h(x) approach 0. To do this we use our knowledge of the sine function. Because the sine of any number lies between and , we can write ≤ sin(1 x) ≤ . Any inequality remains true when multiplied by a positive number. We know that x8 ≥ 0 for all x and so, multiplying each side of inequalities by x8, we get ≤ x8 sin(1 x) ≤ as illustrated by the figure. We know that limx→0 x8 = and limx→0(−x8) = . Taking f(x) = −x8, g(x) = x8 sin(1 /x), and h(x) = x8 in the Squeeze Theorem, we obtain limx→0 x8 sin(1 x) = 0
Suppose that the graph of y = log2(x) is drawn on a coordinate grid where the unit of measurement is an inch. How many miles to the right of the origin do we have to move before the height of the curve reaches 2.5 ft? (Round your answer to one decimal place.) mi
If a bacteria population starts with 140 bacteria and doubles every two hours, then the number of bacteria after t hours is n = f(t) = 140⋅2t/2. (a) Find the inverse of this function. f−1(n) = Explain its meaning. This function tells us how long it will take to obtain n bacteria (given the number n). This function tells us how long it will take to obtain n bacteria (given the number t). (b) When will the population reach 10, 000? (Round your answer to one decimal place.) hr
A sample of a radioactive substance decayed to 92.5% of its original amount after a year. (Round your answers to two decimal places.) (a) What is the half-life of the substance? yr (b) How long would it take the sample to decay to 10% of its original amount? yr
Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14C, with a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore, the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 64% as much 14C radioactivity as does plant material on Earth today. Estimate the age of the parchment. (Round your answer to the nearest hundred years.) yr
Dinosaur fossils are too old to be reliably dated using carbon-14, which has a half-life of about 5730 years. Suppose we had a 69 million year old dinosaur fossil. How much of the living dinosaur's 14C would be remaining today? (Round your answer to five decimal places.) % Suppose the minimum detectable amount is 0.3%. What is the maximum age of a fossil that we could date using 14C? (Round your answer to the nearest integer.) yr
A tank holds 3000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume V of water remaining in the tank (in gallons) after t minutes. (a) If P is the point (15, 765) on the graph of V, find the slopes of the secant lines PQ when Q is the point on the graph with the following values. (Round your answers to one decimal place.) Q (5, 2082)(10, 1338)(20, 351)(25, 81)(30, 0) slope (b) Estimate the slope of the tangent line at P by averaging the slopes of two adjacent secant lines. (Round your answer to one decimal place.)
A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with the given values of t. (Round your answers to one decimal place.) (a) t = 36 and t = 42 (b) t = 38 and t = 42 (c) t = 40 and t = 42 (d) t = 42 and t = 44
The point P(5, −4) lies on the curve y = 4/(4−x). (a) If Q is the point (x, 4 /(4−x)), use your calculator to find the slope mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 4.9 mPQ = (ii) 4.99 mPQ = (iii) 4.999 mPQ = (iv) 4.9999 mPQ = (v) 5.1 mPQ = (vi) 5.01 mPQ = (vii) 5.001 mPQ = (viii) 5.0001 mPQ = (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(5, −4). m = (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(5, −4).
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet t seconds later is given by y = 40t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds ft/s (ii) 0.1 seconds ft/s (iii) 0.05 seconds ft/s (iv) 0.01 seconds ft/s (b) Estimate the instantaneous velocity when t = 2. ft/s
If an arrow is shot upward on Mars with a speed of 54 m/s, its height in meters t seconds later is given by y = 54t − 1.86t2. (Round your answers to two decimal places.) (a) Find the average speed over the given time intervals. (i) [1, 2] m/s (ii) [1, 1.5] m/s (iii) [1, 1.1] m/s (iv) [1, 1.01] m/s (v) [1, 1.001] m/s (b) Estimate the speed when t = 1. m/s
Explain the meaning of each of the following. (a) limx→−4 f(x) = ∞ The values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) -4. The values of f(x) can be made arbitrarily close to -4 by taking x sufficiently large. The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) -4. f(−4) = ∞ (b) limx→9+f(x) = −∞ The values of f(x) can be made arbitrarily close to −∞ by taking x sufficiently close to 9. As x approaches 9, f(x) approaches −∞. f(9) = −∞ The values of f(x) can be made negative with arbitrarily large absolute values by taking x sufficiently close to, but greater than, 9.
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) limx→2−f(x) (b) limx→2+f(x) (c) limx→2 f(x) (d) f(2) (e) limx→4 f(x) (f) f(4)
For the function f whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) (a) limx→1 f(x) (b) limx→3−f(x) (c) limx→3+f(x) (d) limx→3 f(x) (e) f(3)
For the function g whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.) (a) limt→0−g(t) (b) limt→0+g(t) (c) limt→0 g(t) (d) limt→2−g(t) (e) limt→2+g(t) (f) limt→2 g(t) (g) g(2) (h) limt→4 g(t)
A patient receives a 150-mg injection of a drug every 4 hours. The graph shows the amount f(t) of the drug in the bloodstream after t hours. Find limt→4− f(t) and limt→4+ f(t). limt→4−f(t) = mglimt→4+ f(t) = mg
Sketch the graph of an example of a function f that satisfies all of the given conditions. limx→0−f(x) = −1, limx→0+f(x) = 2, f(0) = 1
Sketch the graph of an example of a function f that satisfies all of the given conditions. limx→4+f(x) = 5, limx→4−f(x) = 3, limx→−1 f(x) = 3, f(4) = 4, f(−1) = 2
The graphs of f and g are given. Use them to evaluate each limit, if it exists. (If an answer does not exist, enter DNE.) (a) limx→2[f(x) + g(x)] (b) limx→0[f(x) − g(x)] (c) limx→−1[f(x)g(x)] (d) limx→3 f(x) g(x) (e) limx→2[x2 f(x)] (f) f(−1) + limx→−1 g(x)
(a) What is wrong with the following equation? x2+x−42 x−6 = x+7 (x−6)(x+7) ≠ x2+x−42 The left-hand side is not defined for x = 0, but the right-hand side is. The left-hand side is not defined for x = 6, but the right-hand side is. None of these - the equation is correct. (b) In view of part (a), explain why the equation limx→6 x2+x−42 x−6 = limx→6 (x+7) is correct. Since x2+x−42 x−6 and x+7 are both continuous, the equation follows. Since the equation holds for all x≠6, it follows that both sides of the equation approach the same limit as x→6. This equation follows from the fact that the equation in part (a) is correct. None of these - the equation is not correct.
If limx→0 f(x) x2 = 2, evaluate the following limits. (a) limx→0 f(x) (b) limx→0 f(x) x Find the number a such that the limit exists. a = limx→−2 2x2 + ax + a + 6 x2 + x − 2 Find the value of the limit.
From the graph of f, state each x-value at which f is discontinuous. For each x-value, determine whether f is continuous from the right, or from the left, or neither. (Enter your answers from smallest to largest.) x = (smallest value) continuous from the right continuous from the left neither x = continuous from the right continuous from the left neither x = continuous from the right continuous from the left neither x = (largest value) continuous from the right continuous from the left
Sketch the graph of a function f that is continuous except for the stated discontinuity. Discontinuous at 2, but continuous from the right there
Sketch the graph of a function f that is continuous except for the stated discontinuity. Removable discontinuity at 2, jump discontinuity at 4
Which of the following are continuous functions? (Select all that apply.) The temperature at a specific location as a function of time. The temperature at a specific time as a function of the distance due west from New York City. The altitude above sea level as a function of the distance due west from New York City. The cost of a taxi ride as a function of the distance traveled. The current in the circuit for the lights in a room as a function of time. None of these.
How would you "remove the discontinuity" of f? In other words, how would you define f(3) in order to make f continuous at 3? f(x) = x2−x−6 x−3 f(3) =
Explain, using the theorems, why the function is continuous at every number in its domain. F(x) = 2x2−x−1 x2+1 F(x) is a polynomial, so it is continuous at every number in its domain. F(x) is a rational function, so it is continuous at every number in its domain. F(x) is a composition of functions that are continuous for all real numbers, so it is continuous at every number in its domain. F(x) is not continuous at every number in its domain. none of these State the domain. (Enter your answer using interval notation.)
Explain, using the theorems, why the function is continuous at every number in its domain. Q(x) = x−73 x3−7 Q(x) is a polynomial, so it is continuous at every number in its domain. Q(x) is a rational function, so it is continuous at every number in its domain. Q(x) is a composition of functions that are continuous for all real numbers, so it is continuous at every number in its domain. Q(x) is not continuous at every number in its domain. none of these State the domain. (Enter your answer using interval notation.)
Explain, using the theorems, why the function is continuous at every number in its domain. M(x) = 1 + 2 x M(x) is a polynomial, so it is continuous at every number in its domain. M(x) is a rational function, so it is continuous at every number in its domain. M(x) is a composition of functions that are continuous, so it is continuous at every number in its domain. M(x) is not continuous at every number in its domain. none of these State the domain. (Enter your answer using interval notation.)
Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from the left, or neither. f(x) = {x + 4 if x < 0 4x2 if 0 ≤ x ≤ 1 4 − x if x > 1 x = (smaller value) continuous from the right continuous from the left neither x = (larger value) continuous from the right continuous from the left neither Sketch the graph of f.
The gravitational force exerted by the planet Earth on a unit mass at a distance r from the center of the planet is F(r) = {GMr R3 if r < R GM r2 if r ≥ R where M is the mass of Earth, R is its radius, and G is the gravitational constant. Is F a continuous function of r? Yes, F is a continuous function of r. No, F is not a continuous function of r
For what value of the constant c is the function f continuous on (−∞, ∞)? f(x) = {cx2 + 6x if x < 5 x3 − cx if x ≥ 5 c =
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. x4+x−9 = 0, (1, 2) f(x) = x4+x−9 is on the closed interval [1, 2], f(1) = , and f(2) = . Since −7 < < 9, there is a number c in (1, 2) such that f(c) = ? by the Intermediate Value Theorem. Thus, there is a of the equation x4+x−9 = 0 in the interval (1, 2).