CHAPTER 1 LIMITS
SECTION 1.6 Continuity of Functions
Problem Set 1.6, Number 36 – 71.
- Sketch the graph of a function f that satisfies all the following conditions.
- Sketch the graph of a function that has domain [0, 2] and is continuous on [0, 2) but not on [0, 2].
- Sketch the graph of a function that has domain [0, 6] and is continuous on [0, 2] and (2, 6] but is not continuous on [0, 6].
- Sketch the graph of a function that has domain [0, 6] and is continuous on (0,6) but not on [0, 6].
- Let
- f(x) = sin x; c = 0
- f(x) =x2 – 100; c = 10x – 10
- f(x) =sin x; c = 0x
- f(x) =cos x: c = 0x
- g(x)={sin x, x ≠ 0x0, x = 0
- F(x) =x sin1; c = 0x
- f(x) =sin1; c = 0x
- f(x) =4 – x; c = 42 – √x
- A cell phone company charges $0.12 for connecting a call plus $0.08 per minute or any part thereof (e.g., a phone call lasting 2 minutes and 5 seconds costs $0.12 + 3 × $0.08). Sketch a graph of the cost of making a call as a function of the length of timet that the call lasts. Discuss the continuity of this function.
- A rental car company charges $20 for one day, allowing up to 200 miles. For each additional 100 miles, or any fraction thereof, the company charges $18. Sketch a graph of the cost for renting a car for one day as a function of the miles driven. Discuss the continuity of this function.
- A cab company charges $2.50 for the first 1/4 mile and $0.20 for each additional 1/8 mile. Sketch a graph of the cost of a cab ride as a function of the number of miles driven. Discuss the continuity of this function.
- Use the Intermediate Value Theorem to prove that x3 + 3x – 2 = 0 has a real solution between 0 and 1.
- Use the Intermediate Value Theorem to prove that (cos t) t3 + 6 sin5 t – 3 = 0 has a real solution between 0 and 2π.
- Use the Intermediate Value Theorem to show that x3 – 7x2 + 14x – 8 = 0 has at least one solution in the interval [0, 5]. Sketch the graph of y = x3 – 7x2 + 14x – 8 over [0, 5]. How many solutions does this equation really have?
- Use the Intermediate Value Theorem to show that √x – cos x = 0 has a solution between 0 and π/2. Zoom in on the graph of y = √x – cos x to find an interval having length 0.1 that contains this solution.
- Show that the equation x5 + 4x3 – 7x + 14 = 0 has at least one real solution.
- Prove that f is continuous at c if and only if limt→0 f(c + t) = f(c).
- Prove that if f is continuous at c and f(c) > 0 there is an interval (c – δ, c + δ) such that f(x) > 0 on this interval.
- Prove that if f is continuous on [0, 1] and satisfies 0 ≤ f(x) ≤ 1 there, then f has a fixed point; that is, there is a number c in [0, 1] such that f(c) = c. Hint: Apply the Intermediate Value Theorem to g(x) = x – f(x).
- Find the values of a and b so that the following function is continuous everywhere.
- A stretched elastic string covers the interval [0, 1]. The ends are released and the string contracts so that it covers the interval [a, b] , a ≥ 0, b ≤ 1 Prove that this results in at least one point of the string being where it was originally. See Problem 59.
- Let f(x) = 1/x – 1. Then f(–2) = –1/3 and f(2) = 1. Does the Intermediate Value Theorem imply the existence of a number c between –2 and 2 such that f(c) = 0? Explain.
- Starting at 4 A.M., a hiker slowly climbed to the top of a mountain, arriving at noon. The next day, he returned along the same path, starting at 5 A.M. and getting to the bottom at 11 Α.Μ. Show that at some point along the path his watch showed the same time on both days.
- Let D be a bounded, but otherwise arbitrary, region in the first quadrant. Given an angle θ, 0 ≤ θ ≤ π/2 , D can be circum-scribed by a rectangle whose base makes angle θ with the x-axis as shown in Figure 15. Prove that at some angle this rectangle is a square. (This means that any bounded region can be circum-scribed by a square.)
- The gravitational force exerted by the earth on an object having mass m that is a distance r from the center of the earth is
- Suppose that f is continuous on [a, b] and it is never zero there. Is it possible that f changes sign on [a, b] ? Explain.
- Let f(x + y) = f(x) + f(y) for all x and y and suppose that fis continuous at x = 0.
- Prove that if f(x) is a continuous function on an interval then so is the function |f(x)| = √((f(x))2).
- Show that if g(x) = |f(x)| is continuous it is not necessarily true that f(x) is continuous.
- Let f(x) = 0 if x is irrational and let f(x) = 1/q if x is the rational number p/q in reduced form (q > 0) .
- A thin equilateral triangular block of side length 1 unit has its face in the vertical xy-plane with a vertex V at the origin. Under the influence of gravity, it will rotate about V until a side hits the x-axis floor (Figure 16). Let x denote the initial x-coordinate of the midpoint M of the side opposite V, and let f(x) denote the final x-coordinate of this point. Assume that the block balances when M is directly above V.
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(a)
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Its domain is [–2,
2].
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(b)
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f(–2) = f(–1)
= f(1) = f(2) = 1.
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(c)
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It is
discontinuous at –1 and 1.
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(d)
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It is right
continuous at –1 and left continuous at 1.
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❤
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PEMBAHASAN:
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❤
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PEMBAHASAN:
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❤
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PEMBAHASAN:
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❤
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PEMBAHASAN:
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f(x)
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=
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{
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x
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if x is rational
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–x
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if x is irrational
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Sketch the graph of this function as best you can and decide where it is continuous.
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❤
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PEMBAHASAN:
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Discontinuous at all points except x = 0, because limx→c ≠ f(c) for c ≠ 0.
limx→c f(x) exists only at c = 0 and limx→c f(x) = 0 = f(0).
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BACA JUGA:
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In Problems 41–48, determine whether the function is continuous at the given point c. If the function is not continuous, determine whether the discontinuity is removable or nonremovable.
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❤
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PEMBAHASAN:
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Continuous.
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❤
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PEMBAHASAN:
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Discontinuous: removable, define f(10) = 20.
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❤
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PEMBAHASAN:
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Discontinuous: removable, define f(0) = 1.
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❤
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PEMBAHASAN:
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Discontinuous: nonremovable.
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❤
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PEMBAHASAN:
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Discontinuous, removable, redefine g(0) = 1.
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❤
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PEMBAHASAN:
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Discontinuous: removable, define F(0) = 0.
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❤
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PEMBAHASAN:
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Discontinuous: nonremovable.
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❤
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PEMBAHASAN:
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Discontinuous: removable, define f(4) = 4
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❤
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PEMBAHASAN:
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The function is continuous on the intervals (0,1], (1,2], (2,3], …
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❤
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PEMBAHASAN:
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The function is continuous on the intervals [0,200], (200,300], (300,400], …
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BACA JUGA:
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❤
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PEMBAHASAN:
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The function is continuous on the intervals (0,0.25], (0.25,0.375], (0.375,0.5], …
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❤
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PEMBAHASAN:
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Let f(x) = x3 + 2x − 2. f is continuous on [0, 1]. f(0) = –2 < 0 and f(1) = 2 > 0. Thus, there is at least one number c between 0 and 1 such that x3 + 3x – 2 = 0.
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❤
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PEMBAHASAN:
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Because the function is continuous on [0,2π] and (cos 0)03 + 6sin5 0 – 3 = –3 < 0, (cos 2π)(2π)3 + 6sin5 (2π) 2π) – 3 = 8π3 – 3 > 0, there is at least one number c between 0 and 2π such that (cos t)t)3 + 6sin5 t – 3 = 0.
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❤
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PEMBAHASAN:
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Let f(x) = x3 – 7x2 + 14x – 8. f(x) is continuous at all values of x. f(0) = –8, f(5) = 12. Because 0 is between –8 and 12, there is at least one number c between 0 and 5 such that f(x) = x3 – 7x2 + 14x – 8 = 0.
This equation has three solutions (x = 1,2,4).
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❤
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PEMBAHASAN:
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Let f(x) = √x − cos x. f(x) is continuous at all values of (x) ≥ 0. f(0) = –1, f(π/2) = √(π/2). Because 0 is between –1 and √(π/2), there is at least one number c between 0 and π/2 such that f(x) = √x − cos x = 0.
The interval [0.6,0.7] contains the solution.
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❤
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PEMBAHASAN:
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Let f(x) = x5 + 4x3 – 7x + 14. f(x) is continuous at all values of x. f(–2) = –36, f(0) = 14. Because 0 is between –36 and 14, there is at least one number c between –2 and 0 such that f(x) = x5 + 4x3 – 7x + 14 = 0.
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❤
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PEMBAHASAN:
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Suppose that f is continuous at c, so limx→c f(x) = f(c). Let x = c + t, so t = x – c, then as x→c, t→0 and the statement limx→c f(x) = f(c) becomes limt→0 f(x + c) = f(c).
Suppose that limt→0 f(x + c) = f(c) and let x = t + c, so t = x – c. Since c is fixed, t→0 means that x→c and the statement limt→0 f(x + c) = f(c) becomes limx→c f(x) = f(c), so f is continuous at c.
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❤
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PEMBAHASAN:
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Since f(x) is continuous at c, lim x→c f(x) = f(c) > 0.
Choose ε = f(c), then there exists a δ > 0 such that 0 < |x – c| < δ ⇒ |f(x) – f(c)| < ε.
Thus, f(x) – f(c) > –ε = –f(c), or f(x) > 0.
Since also f(c) > 0, f(x) > 0 for all x in (с – δ, с + δ).
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❤
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PEMBAHASAN:
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Let g(x) = x – f(x). Then, g(0) = 0 – f(0) = –f(0) ≤ 0 and g(1) = 1 – f(1) ≥ 0 since 0 ≤ f(x) ≤ 1 on [0, 1].
If g(0) = 0, then f(0) = 0 and c = 0 is a fixed point of f.
If g(1) = 0, then f(1) = 1 and c = 1 is a fixed point of f.
If neither g(0) = 0 nor g(1) = 0, then g(0) < 0 and g(1) > 0 so there is some c in [0, 1] such that g(c) = 0. If g(c) = 0 then c – f(c) = 0 or f(c) = c and c is a fixed point of f.
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f(x)
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=
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{
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x + 1
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if
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x < 0
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ax + b
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if
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0 ≤ x <
2
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|||
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3x
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if
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x ≥ 2
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❤
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PEMBAHASAN:
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For f(x) to be continuous everywhere,
f(1) = a(1) + b = 2 and f(2) = 6 = a(2) + b
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a + b
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=
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2
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2a + b
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=
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6
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–
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–a
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=
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–4
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a
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=
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4
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b
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=
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–2
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BACA JUGA:
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❤
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PEMBAHASAN:
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For x in [0, 1], let f(x) indicate where the string originally at x ends up.
Thus f(0) = a, f(1) = b.
f(x) is continuous since the string is unbroken.
Since 0 ≤ a, b ≤ 1, f(x) satisfies the conditions of Problem 59, so there is some c in [0, 1] with f(c) = c, i.e., the point of string originally at c ends up at c.
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❤
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PEMBAHASAN:
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The Intermediate Value Theorem does not imply the existence of a number c between –2 and 2 such that f(c) = 0. The reason is that the function f(x) is not continuous on [–2, 2].
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❤
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PEMBAHASAN:
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Let f(x) be the difference in times on the hiker’s watch where x is a point on the path, and suppose x = 0 at the bottom and x = 1 at the top of the mountain.
So f(x) = (time on watch on the way up) – (time on watch on the way down).
f(0) = 4 – 11 = –7, f(1) = 12 – 5 = 7. Since time is continuous, f(x) is continuous, hence there is some c between 0 and 1 where f(c) = 0. This c is the point where the hiker’s watch showed the same time on both days.
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❤
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PEMBAHASAN:
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Let f be the function on [0, π/2] such thath f(θ) is the length of the side of the rectangle which makes angle θ with the x-axis minus the length of the sides perpendicular to it.
f is continuous on [0, π/2].
If f(0) = 0 then the region is circumscribed by a square. If f(0) ≠ 0, then observe that f(0) = –f(π/2).
Thus, by the Intermediate Value Theorem, there is an angle θ0 between 0 and π/2 such that f(θ0) = 0.
Hence, D can be circumscribed by a square.
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f(x)
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=
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{
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GMmr
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,
if
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r < R
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R3
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|||||
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GMm
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,
if
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r ≥ R
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|||
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r2
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Here G is the gravitational constant, M is the mass of the earth, and R is the earth's radius. Is g a continuous function of r?
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❤
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PEMBAHASAN:
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Yes, g is continuous at R.
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❤
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PEMBAHASAN:
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No. By the Intermediate Value Theorem, if f were to change signs on [a,b], then f must be 0 at some c in [a,b]. Therefore, f cannot change sign.
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(a)
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Prove
that fis continuous everywhere.
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(b)
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Prove
that there is a constant m such that f(t) = mt
for all t (see Problem 43 of Section 0.5).
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❤
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PEMBAHASAN:
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(a)
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f(x) = f(x
+ 0) = f(x) + f(0), so f(0) = 0. We want to prove
that limx→c f(x) = f(c),
or, equivalently, limx→c [f(x)
– f(c)] = 0.
But f(x) – f(c) = f(x
– c), so limx→c [f(x)
– f(c)] = limx→c f(x
– c).
Let h = x – c then as x→c,
h→0 and limx→c f(x –
c) = limh→0 f(h) = f(0)
= 0.
Hence limx→c f(x)
= f(c) and f is continuous at c.
Thus, f is continuous everywhere, since c
was arbitrary.
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(b)
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By Problem 43 of Section 0.5, f(t) = mt
for all t in Q.
Since g(t) = mt is a polynomial
function, it is continuous for all real numbers.
f(t) = g(t)
for all t in Q, thus f(t) = g(t)
for all t in R, i.e.
f(t) = mt.
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❤
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PEMBAHASAN:
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If f(x) is continuous on an interval then limx→c f(x) = f(c) for all points in the interval:
limx→c f(x) = f(c) ⇒ limx→c |f(x)|
= limx→c √[f2(x)] = √[(limx→c f(x))2]
= √[(f(c))2] = |f(c)|.
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❤
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PEMBAHASAN:
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Suppose
f(x)
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=
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{
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1 if x ≥ 0
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–1 if x < 0
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f(x) discontinuous at x = 0, but g(x) = |f(x)| = 1 is continuous everywhere.
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Sketch
(as best you can) the graph of f on (0, 1).
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(b)
|
Show
that f is continuous at each irrational number in (0, 1), but is
discontinuous at each rational number in (0, 1).
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❤
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PEMBAHASAN:
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(a)
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(b)
|
If r is any rational number, then any deleted
interval about r contains an irrational number.
Thus, if f(r) = 1/q,
any deleted interval about r contains at least one point c such
that |f(r) – f(c)| = |1/q
– 0| = 1/q.
Hence, limx→r f(x)
does not exist.
If c is any irrational number in (0, 1), then as
x = p/q → c (where p/q
is the reduced form of the rational number) q → ∞, so f(x)
→ 0 as x → c.
Thus, f(x) = 0 = f(c) for
any irrational number c.
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Determine
the domain and range of f.
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(b)
|
Where
on this domain is f discontinuous?
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(c)
|
Identify
any fixed points of f (see Problem 59).
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❤
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PEMBAHASAN:
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(a)
|
Suppose the block rotates to the left. Using geometry, f(x)
= –3/4.
Suppose the block rotates to the right. Using geometry,
f(x) = 3/4.
If x = 0, the block does not rotate, so f(x)
= 0.
Domain: [–3/4, 3/4];
Range: {–3/4, 0, 3/4}.
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(b)
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At x = 0
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(c)
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If x = 0, f (x) = 0, if x =
–3/4, f(x) = –3/4
and if x = 3/4, f(x) = 3/4,
so x = –3/4, 0, 3/4 are
fixed points of f.
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BACA JUGA:
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