MATH225 - Calculus Week 6 Test-Grade-100% 2025-2026 reviewer American Public University

MATH225 - Calculus Week 6 Test-Grade: 100% 2025-2026 reviewer American Public University

Week 6 Test

- Grade: 100%

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Questions 20 Questions

Limits 180 Minutes

Points 100 pts possible

Due Date No due date.

Attempt 1

100% (100 of 100)

Completed on 04/09/25 at 07:16PM

Score for this quiz: 100% ( 100 /100) Submitted Apr 9 at 7:16pm This attempt took 37 minutes.

Question 1 : 5 ptsSkip to question text.

Evaluate limx→∞ (x + 2−−−−−√ – x−−√).Evaluate limx→∞ (x + 2 – x). 2 0

∞∞ 2– √2 lim x→∞(x + 2−−−−−−√ – x−−√) = limx→∞ 2x + 2−−−−−−√ + x−−√=2∞=0lim x→∞(x + 2 – x) = limx→∞ 2x + 2 + x=2 ∞=0 5 /5

Question 2 : 5 ptsSkip to question text.

Evaluate limx→2 x3 − 8x4 − 16.Evaluate limx→2 x3 − 8x4 − 16. 1/2 −3/8 3/4 3/8

limx→2 x3 − 8x4 − 16 = limx→2 3x24x3 (form 0/0, use L’Hôpital’s rule) the values) = 38limx→2 x3 − 8x4 − 16 = limx→2 3x24x3 = 3⋅224⋅23 (plug in the values) = 38

= 3⋅224⋅23 (plug in (form 0/0, use L’Hôpital’s rule)

5 /5 Question 3 : 5 ptsA car is moving at a constant acceleration of 3 m / sec2. If the car is moving with a velocity of 20 m / sec at t = 0, how fast is the car moving when t = 3?

A car is moving at a constant acceleration of 3 m / sec2. If the car is moving with a velocity of 20 m / sec at t = 0, how fast is the car moving when t = 3?

32 m / sec 29 m / sec 23 m / sec 9 m / sec

Since the car is moving with a constant acceleration of 3 m /sec2, then the velocity of the car will increase by 3 m /sec every second, or 9 m /sec over 3 seconds. So the car will be moving at 29 m /sec when t = 3.

5 /5 Question 4 : 5 ptsSkip to question text.

Evaluate limx → 1 (x−1)3(1−x)−2.Evaluate limx → 1 (x−1)3(1−x)−2.

0 3/2 3 The limit does not exist.

limx → 1 (x−1)3(1−x)−2→0⋅∞limx → 1 (x−1)3(1−x)−2=limx → 1 (x−1)3(1−x)2=limx → 1 3(x−1)2−2(1−x)=limx → 1 6(x−1)2=0limx → 1 (x−1)3(1−x)−2→0⋅∞limx → 1 (x−1)3(1−x)−2=limx → 1 (x−1)3(1−x)2=limx → 1 3(x−1)2−2(1−x)=limx → 1 6(x−1)2=0 5 /5 Question 5 : 5 ptsSkip to question text.

Evaluatelim x → ∞ 4x5 +10x 3+9x 2+2x+12x 5−3x 4−9x+5.Evaluatelimx → ∞ 4x5+10x3+9x2+2x+12x5−3x4−9x+5.

2 −2 1515

−265−265

lim x → ∞ 4x5 +10x3 +9x2 +2x+12x 5 −3x4 −9x+5=lim x → ∞4+10x2+9x3+2x4+1x52−3x−9x4−5x5=4+0+0+0+02−0−0+0=2 limx → ∞4x5+10x3+9x2+2x+12x5−3x4−9x+5=limx → ∞4+10x2+9x3+2x4+1x52−3x−9x4−5x5=4+0+0+0+02−0−0+0=2 5 /5 Question 6 : 5 ptsYour average soda can has a height of 5 inches and a circular top and bottom with a diameter of 2.5 inches. What is the volume of this soda can?

Your average soda can has a height of 5 inches and a circular top and bottom with a diameter of 2.5 inches. What is the volume of this soda can?

2.6π in32.6π in3 7.8125π in37.8125π in3 10.42π in310.42π in3 31.25π in331.25π in3

The formula for the volume of a cylinderis V=πr2h, where r is the radius of thecircular base (top or bottom) and h isth e height of the cylinder.The diameter of the bottom of the cylinderis 2.5 inches. So, the radius is 1.25 inches. Substitu te th e given value of the height, 5 inches, and the radius, 1.25 inches, intothe formula for the volume of a cylinder.V= πr2h=π(1.25)2(5)=7.8125πSo, the volume of an average soda can is 7.8125π in 3.The formula for the volume of a cylind eris V=πr2h, where r is the radius of thecircular base (top or bottom) and h isthe height of the cylinder.The diameter of th e bottom of the cylinderis 2.5 inches. So, the radius is 1.25 inches. Substitute the given value of the height, 5 inches, and t he radius, 1.25 inches, intothe formula for the volume of a cylinder.V=πr2h=π(1.25)2(5)=7.8125πSo, the volume of an ave rage soda can is 7.8125π in3.

5 /5 Question 7 : 5 ptsSkip to question text.

A particular company manufactures cardboard boxes of many different volumes. They always start with a flat sheet of cardboard . By cutting an equal sized square from each corner and folding the sides upwards, they make a box. If the cardboard sheet is 18 × 16 inches and you cut a 2 × 2 square from each corner, what would be the volume of the resulting box?

448 cubic inches 392 cubic inches 336 cubic inches 168 cubic inches

Since the cardboard sheet is 18 × 16, subtracting off 2 × 2 square corners and folding up will construct a topless box. But notice that the 2 × 2 squares that you are subtracting are subtracted from all four corners. If the length of the cardboard sheet is 18, then subtracting two 2 × 2 squares would leave you with 14 inches. If the width is 16, then you would be left with 12 inches. So the dimensions of the cardboard box would be 14 × 12 × 2 with a volume of 336 cubic inches.

5 /5 Question 8 : 5 ptsSkip to question text.

A certain jet plane consumes fuel at a rate (in lbs / hour) given by the function F(x) = x4 − 20x3 + 28,000, where x is the amount of fuel additive used in ounces.

What is the minimum rate of fuel consumption of the plane (in lbs / hour)? 15 11,125 28,000 146,125

The minimum rate of fuel consuption will occur when f ′(x) = 0.

So, first find the derivative f ′(x): F(x) = x 4 − 20x 3 + 28,000 f ′(x) = 4x 3 − 60x 2.

Now set f ′(x) = 0 and solve for x: 4x 3 − 60x 2 = 0 x 3 − 15x 2 = 0 x 2 (x − 15) = 0 x = 0 or x = 15.

These points are the candidates for the minimum value of F(x). Plug them in to find the minimum value: F(0) = 0 − 0 + 28,000 F(15) = (15)4 − 20(15)3 + 28,000 = 50,625 − 67,500 + 28,000 = 11,125.

The minimum rate occurs when x = 15, and it is equal to 11,125 lbs /hour.

5 /5 Question 9 : 5 ptsSkip to question text.

Evaluate limx → π/2 sinx+cosx−1cosx.Evaluate limx → π/2 sinx+cosx−1cosx. 0 1 −1 The limit does not exist.

limx → π/2sinx+cosx−1cosx→00limx → π/2sinx+cosx−1cosx=limx → π/2cosx−sinx−sinx=cos(π/2)−sin(π/2)−sin(π/2)=0−1−1=1 limx → π/2sinx+cosx−1cosx→00limx → π/2sinx+cosx−1cosx=limx → π/2cosx−sinx−sinx=cos(π/2)−sin(π/2)−sin(π/2)=0−1−1=1 5 /5

Question 10 : 5 ptsA farmer needs a new pen for some of his livestock and decides to build a rectangular fence. The farmer has 60 feet of fence. What is the greatest area the farmer could enclose?

A farmer needs a new pen for some of his livestock and decides to build a rectangular fence. The farmer has 60 feet of fence. What is the greatest area the farmer could enclose?

0 feet 2 300 feet 2 200 feet 2 225 feet 2

First, figure out what you know. Let b and w be the base and width of the pen and let A and P be the area and perimeter of the pen (respectively).

Then: A=bwP=2b+2w=602b=60−2wb=30−wA=w(30−w)A=bwP=2b+2w=602b=60−2wb=30−wA=w(30−w)

To maximize the area, take the derivative and set it equal to zero. dAdw=d(30w−w2)dw=d(30w)dw−d(w2)dw=30−2w0=30−2w2w=30w=15b=15A=15⋅15=225 =d(30w)dw−d(w2)dw=30−2w0=30−2w2w=30w=15b=15A=15·15=225 ft2

ft2dAdw=d(30w−w2)dw

5 /5 Question 11 : 5 ptsSkip to question text.

Evaluate limx→1x3−2x+1x4+3x−4.Evaluate limx→1x3−2x+1x4+3x−4. 0 1/7 −1/4 1/2

limx→1x3−2x+1x4+3x−4=limx→13x2−24x3+3 =17limx→1x3−2x+1x4+3x−4=limx→13x2−24x3+3

=3⋅1−24⋅1+3 =3⋅1−24⋅1+3

=17

5 /5 Question 12 : 5 ptsSkip to question text.

Suppose the position of an object on a number line is given by p(t) = t2 − 3t + 1 where t is in seconds and p is in centimeters. What is the acceleration of the object when t = 2 seconds?

4 cm / sec2 2 cm / sec2 1 cm / sec2 −1 cm / sec2

p(t)=t2−3t+1p′(t)=2t−3p′′(t)=2p′′(2)=2p(t)=t2−3t+1p′(t)=2t−3p″(t)=2p″(2)=2 5 /5 Question 13 : 5 ptsSkip to question text.

Evaluate limx→3 2x−6x2−4x+3.Evaluate limx→3 2x−6x2−4x+3. 1/2 0 ∞ 1

limx→3 2x − 6x2 − 4x + 3 = limx→3 2(x − 3)(x − 1)(x − 3)= limx→3 2(x − 1)= 22= 1limx→3 2x − 6x2 − 4x + 3 = limx→3 2( x − 3)(x − 1)(x − 3)= limx→3 2(x − 1)= 22= 1 5 /5 Question 14 : 5 ptsSkip to question text.

Evaluate limx→∞2x2−75x2+9x+2.Evaluate limx→∞2x2−75x2+9x+2. −72−72 2525 −25−25 1515 limx→∞ 2x2 − 75x2 + 9x + 2 = limx→∞ 4x10x + 9 (form ∞/∞, use L’Hôpital’s rule)= limx→∞ 410 + 0 (form ∞/∞, reuse L’Hôpital’s rule)= 25limx→∞ 2x2 − 75x2 + 9x + 2 = limx→∞ 4x10x + 9 (form ∞/∞, use L’Hôpital’s rule)= limx→∞ 410 + 0 (form ∞/∞, reuse L’Hôpital’s rule)= 25 5 /5 Question 15 : 5 ptsSkip to question text.

Given that a particular moving object'svelocity is given by the equationv(t)=−32t+110, what is the equation for theobject' s acceleration?Given that a particular moving object'svelocity is given by the equationv(t)=−32t+110, what is the equation for theobject's acceleration? a(t) = 110 a(t) = −110 a(t) = 32 a(t) = −32

Since the derivative of velocity equationequals the acceleration equation, justtake the derivative of the velocity.v(t)= −32t+110v′(t)=a(t)=−32Since the derivative of velocity equationequals the acceleration equation, justtake the derivative of the velocity.v(t)=−32t+110v′(t)=a(t)=−32 5 /5 Question 16 : 5 ptsSkip to question text.

Is the point (0, 4) a minimum point of the function? Why or why not? y=x3+4xxy=x3+4xx No, because the function is not differentiable at the point (0, 4). No, because the function is not defined at the point (0, 4). No, because the function reaches its maximum value at (0, 4).

Yes, because the function cannot pass lower than the point (0, 4).

Notice that the function y=x3+4xxis not defined at x=0. So, (0, 4) is noton the curve and it cannot be a minimum poi nt.

Notice that the function y=x3+4xxis not defined at x=0. So, (0, 4) is noton the curve and it cannot be a minimum poin t.

5 /5 Question 17 : 5 ptsSkip to question text.

Evaluate limx→∞(1+5x)x.Evaluate limx→∞(1+5x)x. 5e e5 e 1

limx→∞(1+5x)x=limx→∞exln(1+5/x)limx→∞xln(1+5x)=limx→∞ln(1+5/x)1/x=limx→∞[1/(1+5/x)] (−5x−2)−x−2=limx→∞51+5/x=5limx→∞exln(1+5/x)=e5limx→∞(1+5x)x=limx→∞exln(1+5/x)limx→∞xln(1+5x)=limx→∞ln(1+5/ x)1/x=limx→∞[1/(1+5/x)](−5x−2)−x−2=limx→∞51+5/x=5limx→∞exln(1+5/x)=e5 5 /5 Question 18 : 5 ptsSkip to question text.

A cartoon cat lights a bomb and tosses itinto the air. The bomb is set to detonate in8 seconds. The position of the bomb isg iven by the function P(t)=−9t2+72twith P(t) in feet and t in seconds.Where is the bomb's highest point? A cartoon cat lights a bomb and tosses itinto the air. The bomb is set to detonate in8 seconds. The position of the bomb isgiven by the function P(t)=−9t2+72 twith P(t) in feet and t in seconds.Where is the bomb's highest point?

144 feet 432 feet 4 seconds 8 seconds

To find when the bomb is at its maximumheight, you must find when the first derivative is equal to zero.P(t)=−9t 2+7 2tP′(t)=−18t+72−18t+72=018t=72t=4But you aren't done yet. The question askswhere the bomb is, not when is the bombhighest. So to find the location of the bomb,plug t=4 back into P(t).P(4)=−9(4) 2+72(4)=−144+288=144So, the highest point of the bomb is 144 feet.To find when the bomb is at its maximumheight, you must find when the first deriva tive is equal to zero.P(t)=−9t2+72tP′(t)=−18t+72−18t+72=018t=72t=4But you aren't done yet. The question askswhere the bomb is, not when is the bombhighest. So to find the location of the bomb,plug t=4 back into P(t).P(4)=−9(4)2+72(4)= −144+288=144So, the highest point of the bomb is 144 feet.

5 /5

Question 19 : 5 ptsSkip to question text.

Evaluate.

limx→ 9x−9x2−11x+18limx→ 9x−9x2−11x+18 1/29 1x−21x−2 1/7 −1/2

limx→9x−9x2−11x+18=limx→912x−11=118−11=17limx→9x−9x2−11x+18=limx→912x−11=118−11=17 5 /5 Question 20 : 5 ptsSkip to question text.

Evaluate limx → 0 sin2xcosx−1.Evaluate limx → 0 sin2xcosx−1. −2 0 1 2

limx → 0 sin2xcosx−1→00limx → 0 sin2xcosx−1=limx → 0 2sinxcosx−sinxlimx → 0 2sinxcosx−sinx→00limx → 0 2sinxcosx−sinx=limx → 0 2cosx−1=−2limx → 0 sin2xcosx−1→00limx → 0 sin2xcosx−1=limx → 0 2sinxcosx−sinxlimx → 0 2sinxcosx−sinx→00limx → 0 2sinxcosx−sinx=limx → 0 2cosx−1=−2 5 /5

Quiz Score: 100 out of 100