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MATH 225 Week 1 Test | MATH225 week 1 Answered 100% -Score for this quiz: 100% ( 100 /100)

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Week 1 Test - Grade: 100% Instructions: Questions Limits Points Due Date 15 Questions 180 Minutes 100 pts possible No due date. Attempt 1 100% (100 of 100) Completed on 03/09/25 at 01:42AM Score for this quiz: 100% ( 100 /100) Submitted Mar 9 at 1:42am This attempt took 40 minutes. Question 1 : 6.65 pts If −x2+x+1≤f(x)≤−x+2 for all x, find limx→1f(x). Enter only the value of the limit in the space provided below. If the answer is not an integer, enter it as a fraction in simplest form. Do not enter a mixed number. If the limit does not exist, enter “does not exist”. Question 2 : 6.65 pts f(x)={x, x1x+2, x≥1Evaluate limx→ 1 + f(x). 3 2 1 The one-sided limit does not exist. The plus-sign in the limit indicates a right-handed limit. For right-handed limits, you only have to consider the domain values greater than the one being approached in the limit. f(x)={x, x1x+2, x≥ 1limx→ 1 + f(x)=limx→ 1 + (x+2)=(1+2)=3 Question 3 : 6.65 pts After jumping out of a plane at t=0, a skydiver's altitude in the air in meters isgiven by the position functionp(t)=−50t2+200t+2350, where t is thetime in seconds. At what time will theparachuter be 100 meters from the ground? t = 9 or t = −5 t = 9.28 t = −477650 t = 9 Question 4 : 6.65 pts What is the change in U.S. population from July 1, 1985 to July 1, 1987? Date Nat'l. Pop. Pop. Change July 1, 1999 272,690,8 13 2,442,810 July 1, 1998 270,248,0 03 2,464,396 July 1, 1997 267,783,6 07 2,555,035 July 1, 1996 265,228,5 72 2,425,296 July 1, 1995 262,803,2 76 2,476,255 July 1, 1994 260,327,0 21 2,544,413 July 1, 1993 257,782,6 08 2,752,909 July 1, 1992 255,029,6 99 2,876,607 July 1, 1991 252,153,0 92 2,688,696 July 1, 1990 249,464,3 96 2,645,166 July 1, 1989 246,819,2 30 2,320,248 July 1, 1988 244,498,9 82 2,210,064 July 1, 1987 242,288,9 18 2,156,031 July 1, 1986 240,132,8 87 2,209,092 July 1, 1985 237,923,7 95 2,098,893 July 1, 1984 235,824,9 02 2,032,908 4,365,123 people 2,156,031 people 2,182,562 people 1,455,041 people In 1985, the U.S. population was equal to 237,923,795. In 1987 the population had risen to 242,288,918. The change in population is equal to the difference of the two figures. Question 5 : 6.65 pts f(x)=4x−32x+5Find limx→−2f(x). −1 −3 −2 −11 Question 6 : 6.65 pts Given limx→cf(x)=2 and limx→cg(x)=4, evaluatelimx→c[2f(x)−g(x)]. 6 −4 4 0 Question 7 : 6.65 pts Use the Squeeze Theorem to find limx→0|x|sin10πx. Enter only the value of the limit in the space provided below. If the answer is not an integer, enter it as a fraction in simplest form. Do not enter a mixed number. If the limit does not exist, enter “does not exist”. Question 8 : 6.65 pts f(x)=6x2−7x+2Find limx→7f(x). 7 3 0 6 Question 9 : 6.65 pts f(x)=x2−14x+50Find limx→7f(x). 1 −1 0 7 Question 10 : 6.65 pts Given limx→2f(x)=3 and limx→2g(x)=2,evaluate limx→23f(x)−g(x)g(x). 7/2 1 3/2 8 Question 11 : 6.65 pts f(x)=xFind limx→1f(x). 0 2 1 −1 Question 12 : 6.65 pts f(x)={2x−3, x1x+1, x1Evaluate limx→ 1 − f(x). 2 0 −1 −5 Question 13 : 6.65 ptsLet f (x) = x 2 + 1. What is the average rate of change of f over the interval (2, 2.1)? Let f (x) = x 2 + 1. What is the average rate of change of f over the interval (2, 2.1)? 5.41 4.1 .41 .1 Question 14 : 6.65 pts f(x)=|x−1|Evaluate limx→ 1 − f(x). −2 2 1 0 The negative sign in the limit indicates a left-handed limit. For left-handed limits, you only have to consider the domain values less than the one being approached in the limit. When working with absolute values, it is a good idea to rewrite the function as a piecewise function so you make sure you apply the absolute value. Question 15 : 6.9 pts Suppose f(x)={x+2, x3x2+1, x3. Is f continuous on the interval [−2, 2]? Yes, f is continuous on the interval [−2, 2]. No, f is not continuous on the interval [−2, 2] because it has a point discontinuity. No, f is not continuous on the interval [−2, 2] because it has a jump discontinuity. No, f is not continuous on the interval [−2, 2] because it has an interval discontinuity. For a function to be continuous on an interval, it must be continuous at each point on that interval.

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Week 1 Test - Grade: 100%
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Questions Limits Points Due Date


15 Questions 180 Minutes 100 pts possible No due date.




Attempt 1 100% (100 of 100) Completed on 03/09/25 at 01:42AM
Score for this quiz: 100% ( 100 /100)
Submitted Mar 9 at 1:42am
This attempt took 40 minutes.

Question 1 : 6.65 ptsSkip to question text.
If −x2+x+1≤f(x)≤−x+2 for all x, find limx→1f(x).

Enter only the value of the limit in the space provided below. If the answer is not an integer, enter it as a fraction in simplest form. Do not
enter a mixed number. If the limit does not exist, enter “does not exist”.
Your Answer: 1
Correct Answer(s):
1
Since −x2+x+1≤f(x)≤−x+2, by the Squeeze Theorem, limx→1−x2+x+1≤limx→1f(x)≤limx→1−x+2 −12+1+1≤limx→1f(x)≤−1+2
1≤limx→1f(x)≤1Thus, limx→1f(x)=1.
6..65

, Question 2 : 6.65 ptsSkip to question text.
f(x)={x, x<1x+2, x≥1Evaluate limx→ 1 + f(x).
3
2
1
The one-sided limit does not exist.
The plus-sign in the limit indicates a right-handed limit. For right-handed limits, you only have to consider the domain values greater than the one
being approached in the limit. f(x)={x, x<1x+2, x≥1limx→ 1 + f(x)=limx→ 1 + (x+2)=(1+2)=3
6..65


Question 3 : 6.65 ptsSkip to question text.
After jumping out of a plane at t=0, a skydiver's altitude in the air in meters isgiven by the position functionp(t)=−50t2+200t+2350, where t
is thetime in seconds. At what time will theparachuter be 100 meters from the ground?
t = 9 or t = −5
t = 9.28
t = −477650
t=9
Set p(t)=100 and solve for t.p(t)=−50t2+200t+2350=100−50t2+200t+2250=0−t2+4t+45=0t2−4t−45=0(t−9)(t+5)=0t=9 or t=−5Since t=−5 doesn't ma
ke sense in thecontext of the problem (the skydiver jumpedout of the plane at t=0), we omit thissolution.
6..65


Question 4 : 6.65 ptsSkip to question text.
What is the change in U.S. population from July 1, 1985 to July 1, 1987?
Nat'l. Pop.
Date
Pop. Change
July 1, 272,690,8
2,442,810
1999 13
July 1, 270,248,0
2,464,396
1998 03
July 1, 267,783,6
2,555,035
1997 07
July 1, 265,228,5
2,425,296
1996 72
July 1, 262,803,2
2,476,255
1995 76
July 1, 260,327,0
2,544,413
1994 21
July 1, 257,782,6
2,752,909
1993 08

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