MATH225 - Calculus Week 1 Test -Grade 100% 2025-2026 reviewer American Public University

MATH225 - Calculus Week 1 Test -Grade 100% 2025-2026 reviewer American Public University

Week 1 Test

- Grade: 100%

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Questions 15 Questions

Limits 180 Minutes

Attempt 1

Points

Due Date

100 pts possible

No due date.

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 1≤ limx→1f(x)≤ 1Thus, limx→1f(x)=1.

6.65 / 6.65

−12+1+1≤ limx→1f(x)≤ −1+2

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 / 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+20 0t+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 / 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, 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.

242,288,918 − 237,923,795 = 4,365,123 6.65 / 6.65 Question 5 : 6.65 ptsSkip to question text.

f(x)=4x−32x+5Find limx→−2f(x). −1 −3 −2 −11 None of these

The limit of a rational function as x approaches −2 is the function evaluated at −2. Thus, the limit is (4(−2) − 3) /(2(−2) + 5), or −11.

6.65 / 6.65 Question 6 : 6.65 ptsSkip to question text.

Given limx→cf(x)=2 and limx→cg(x)=4, evaluatelimx→c[2f(x)−g(x)].

6 −4 4 0

limx→cf(x)=2 and limx→cg(x)=4limx→c[2f(x)−g(x)]=limx→c[2f(x)]−limx→cg(x)=2⋅limx→cf(x)−limx→cg(x)=2(2)−4=0 6.65 / 6.65 Question 7 : 6.65 ptsSkip to question text.

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”.

Your Answer: 0 Correct Answer(s): 0

Since −1≤ sin10πx≤ 1, it followsthat −|x|≤ |x|sin10πx≤ |x|. By theSqueeze Theorem, limx→0−|x|≤ limx→0|x|sin10πx≤ limx→0|x|, or0≤ limx→0|x| sin10πx≤ 0. Thus, limx→0|x|sin10πx=0.

6.65 / 6.65 Question 8 : 6.65 ptsSkip to question text.

f(x)=6x2−7x+2Find limx→7f(x).

7 3 0 6 None of these

The limit of a rational function as x approaches 7 is the rational function evaluated at 7 (as long as the denominator is not 0). So the limit is 6 /(72 − 7 (7) + 2), or 3.

6.65 / 6.65 Question 9 : 6.65 ptsSkip to question text.

f(x)=x2−14x+50Find limx→7f(x).

1 −1 0 7 None of these

The limit of the square root of a function as x approaches 7 is the square root of the limit of the function as x approaches 7. Thus, the limit is the square root of 72 − 14(7) + 50, or 1.

6.65 / 6.65

Question 10 : 6.65 ptsSkip to question text.

Given limx→2f(x)=3 and limx→2g(x)=2,evaluate limx→23f(x)−g(x)g(x).

7/2 1 3/2 8

limx→2f(x)=3 and limx→2g(x)=2limx→23f(x)−g(x)g(x)=limx→2[3f(x)−g(x)]limx→2g(x)=limx→2[3f(x)]−limx→2g(x)limx→2g(x)=3⋅limx→2f(x) −limx→2g(x)limx→2g(x)=3⋅(3)−(2)(2)=72 6.65 / 6.65 Question 11 : 6.65 ptsSkip to question text.

f(x)=xFind limx→1f(x).

0 2 1 −1 None of these

The limit of x as x approaches 1 is the value of x at 1. In other words, the limit is 1.

6.65 / 6.65 Question 12 : 6.65 ptsSkip to question text. f(x)={2x−3,

x<1x+1, x>1Evaluate limx→ 1 − f(x). 2 0 −1 −5 None of these

f(x)={2x−3, x<1x+1, x>1limx→ 1 − f(x)=limx→ 1 − (2x−3)=(2−3)=−1 6.65 / 6.65 Question 13 : 6.65 ptsLet f (x) = x2 + 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

f(2.1)=5.41, f(2)=5ΔfΔx=f(2.1)−f(2)2.1−2=5.41−52.1−2=.41.1=4.1 6.65 / 6.65 Question 14 : 6.65 ptsSkip to question text.

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. f(x)=|x−1|={x−1, x≥ 11−x, x<1limx→ 1 − f(x)=limx→ 1 − (1−x)=1−1=0 6.65 / 6.65 Question 15 : 6.9 ptsSkip to question text.

Suppose f(x)={x+2, x<3x2+1, x>3.

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.

Since the function is a piecewise-defined function, you only have to check that the pieces are continuous and that the function is continuous when it changes pieces. But notice that the function is defined by the first piece for the entire given interval. Since lines are continuous, then the function is continuous on this interval.

6.9 / 6.9