MATH225 - Calculus Week 5 Test-Grade- 100% 2026 COMPLETE UPDATE American Public University

MATH225 - Calculus Week 5 Test-Grade- 100% 2026 COMPLETE UPDATE American Public University

Week 5 Test - Grade: 100% Instructions: Click on “Start” to begin the Test.

This Test may be printed by clicking the Print icon at the top of the Test window AFTER starting the Test.

We suggest you work out the answers on the printed Test, then submit your answers online.

THIS IS A TIMED TEST. YOU HAVE 3 HOURS TO COMPLETE THE TEST ONCE YOU CLICK "START." You can start and stop the Test if you need to; however, the time will continue to elapse. You can also skip questions and go back to them as needed during the test. Use the 'skip' button to skip a question and question navigation pull-down menu to jump back to any questions you skipped.

Once you have completed the Test online, click “Submit Answers.” Your answers will be scored and the answer key with step-by-step solutions will become available.

Questions? Reach out to us at support@thinkwell.com. We’re here and happy to help.

Questions 20 Questions

Limits 180 Minutes

Points

Due Date

100 pts possible

No due date.

Attempt 1

65% (65 of 100)

Completed on 04/03/25 at 07:48PM

Attempt 2

100% (100 of 100)

Completed on 04/09/25 at 03:13PM

Score for this quiz: 100% ( 100 /100) Submitted Apr 9 at 3:13pm This attempt took about 1 hour.

Question 1 : 5 ptsSkip to question text.

Below is a graph which has been divided into four sections. In which of these sections is the derivative of the function always negative?

Section A Section B Section C Section D Section E

In section E, the function is increasing, so the derivative there is always positive. In sections A, B and C the function both increases and decreases, so the derivative is positive in some places on the interval and negative in some others. In section D, the function is only decreasing, however, so there the derivative is always negative.

5/5 Question 2 : 5 ptsSkip to question text.

Which of the following curves is the graph of the equation f(x)=x3−x?f(x)=x3−x?

Since f (x) is a rational function, you need to check for asymptotes. Start by looking for vertical asymptotes by setting the denominator equal to zero.

You will find a vertical asymptote at x = 3.

Next look for horizontal asymptotes by taking the limit as the function approaches + /- infinity.

li m

x

= 3 −1 x→ − x ∞ li m

x

= 3 −1 x→− − x ∞

So there is a horizontal asymptote at y = −1.

Next, if you take the first derivative you can determine where the function is increasing and where it is decreasing. The derivative is Notice that the only critical point is our vertical asymptote and that the derivat3ive is always positive. So the function is always increasing. Plotting a f ′ (x) few points will show that (c) is the best answer choice. .

(3 = 5/5 − x)2 Question 3 : 5 ptsSkip to question text.

Which of the following curves is the bestsketch of f(x)=x 2−1−−−−−√?Which of the following curves is the bestsketch of f(x)=x2− 1?

When sketching curves, start by checking the domain for holes. There is an even root in this equation, so you must restrict the expression underneath the radical away from being negative. This means that x must be less than or equal to −1 or greater than or equal to +1.

Now that the domain is properly restricted, look for critical points. To do this you will need to take the first derivative. f′(x)=x(x2−1)12f′(x)=x(x2−1)12 Now that you know the first derivative you can find the critical points by determining where the derivative is undefined or equals zero. Notice that the

derivative must be undefined wherever the function is undefined, but those points are not important since they cannot be critical points. However, the derivative is undefined at x = −1 and x = +1. Now you can use the first derivative test.

Since the curve was decreasing to the left at x = −1, then that point will be a minimum. Since the curve is increasing to the right at x = +1, that point will be also be a minimum.

Now that you have found the critical values, next you must check for concavity and points of inflection. Do this by taking the second derivative. f′′(x)=−x2(x2−1)32+1(x2−1)12f″(x)=−x2(x2−1)32+1(x2−1)12 The second derivative is never equal to zero, but it is undefined at x equals −1 and +1. Check the values of the second derivative around these points to see if either is a point of inflection.

Since the domain is undefined between −1 and +1 you don’t have to check the concavity there. Everywhere else the second derivative is always negative, so the curve is always concave down.

So the curve in question has minimum points at −1 and +1 and is always concave down. In addition, the curve is undefined between −1 and 1. So only answer choice B fits the given function.

5/5 Question 4 : 5 ptsWhat is the minimum value that f (x) = 5x 2 + 10x − 4 attains?

What is the minimum value that f (x) = 5x 2 + 10x − 4 attains?

1 11 −9 −19

The derivative isf′(x)=10x+10=10(x+1)Because the derivative exists everywhere, the only criticalpoint is where the d erivativeequals zero, which occurs atx=−1. The first derivative test tellsyou that this point is a minimum. Plugging x =−1 into the functionexpression producesf(−1)=5(−1) 2+10(−1)−4=−9The derivative isf′(x)=10x+10=10(x+1)Because the derivative exists everywhere, the only criticalpoint is where the derivativeequals zero, which occurs atx=−1. The first derivative test tellsyou that this point is a minimum. Plugging x=−1 into the functionexpression producesf(−1)=5(−1)2+1 0(−1)−4=−9 5/5 Question 5 : 5 ptsSkip to question text.

Find all of the critical points of the function f(x)=15x 5−34x4+23x3and identify them as maxima, minima, or neither. Find all of the critical points of the function f(x)=15x5−34x4+23x3and identify them as maxima, minima, or neither. x = 1 is neither a maximum nor a minimum; x = 0 is a maximum; x = 2 is a minimum x = 2 is neither a maximum nor a minimum; x = 1 is a maximum; x = 0 is a minimum x = 0 is neither a maximum nor a minimum; x = 1 is a maximum; x = 2 is a minimum x = 0 is neither a maximum nor a minimum; x = 2 is a maximum; x = 1 is a minimum

The derivative is f′( x)=x4−3x3+2x2=x2( x−1)( x−2).Because this derivative is defined everywhere, any critical points occurwhere the derivative equals zero.This happens at x=0, x=1 and x=2. The first derivative test tells you that the point where x=1 is the location of a maximum, the point where x=2 is the location of a minimum, and the point wher e x=0 is neither.The derivative is f′( x)=x4−3x3+2x2=x2( x−1)( x−2).Because this derivative is defined everywhere, any c ritical points occurwhere the derivative equals zero.This happens at x=0, x=1 and x=2. The first derivative test tells you that the point where x=1 is the location of a maximum, the point where x=2 is the location of a minimum, and the point w here x=0 is neither.

5/5

Question 6 : 5 ptsSkip to question text.

Given that the function f (x) has a critical point at x = −1 and the second derivative is f ″(−1) = −3, what can be said about f (x) at x = −1?

Nothing can be said about the curve at that point.

There is a point of inflection there.

There is a minimum point there.

There is a maximum point there.

The second derivative test tells us that if the second derivative is negative at a critical point, then the point is a maximum. If the second derivative were positive, then the point would be a minimum.

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

Which of the following is the graph ofthe equation f(x)=x5/3−4x2/3?Which of the following is the graph ofthe equation f(x)=x5/3−4x2/3?

Start by finding the first and second derivatives. f(x)=x2/3(x−4)f '(x)=53x−83x1/3, x≠0f ' '(x)=109+89x−1x1/3, x≠0f(x)=x2/3(x−4)f '(x)=53x−83x1/3, x≠0f ' '(x)=109+89x−1x1/3, x≠0 Now find the critical points. Critical points are located at x-values where the first derivative is equal to zero or undefined. The first derivative is undefined at x = 0 and is equal to zero at 8/5.

After finding the critical points, check the intervals between the points to determine where the curve is increasing and decreasing and to see if the critical points are actually extreme points.

Now you can check for concavity and inflection points. x-values where the second derivative equals zero are inflection point candidates. Also look for places where the second derivative is undefined, since the concavity can change there as well.

The second derivative is equal to zero at x = −4/5 and is undefined at x = 0. Checking the divided intervals will show that the second derivative is negative for values less than −4/5, positive between −4/5 and 0, and positive for values greater than zero. So the graph is concave down for x < −4/5 and concave up for x > −4/5. This also means that there is a point of inflection at −4/5.

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

Use the graph to find the absolute and local maximum and minimum values of the function.

Absolute maximum value: none Absolute minimum value: none Local maximum values: −1 and 5 Local minimum values: −8 and −6 Absolute maximum value: 5 Absolute minimum value: −8 Local maximum values: −1, 5, and 6 Local minimum values: −8, −6, and −5 Absolute maximum value: 5 Absolute minimum value: −8 Local maximum values: −1 and 5 Local minimum values: −8 and −6 Absolute maximum value: 6 Absolute minimum value: −8 Local maximum values: −1, 5, and 6 Local minimum values: −8, −6 and −5

The absolute maximum value is the highest value that the function achieves. The function has no absolute maximum value.

The absolute minimum value is the lowest value that the function achieves. The function has no absolute minimum value.

The local maximum values correspond to the local peaks on the graph. The values of the function at the peaks are −1 and 5.

The local minimum values correspond to the local valleys on the graph. The values of the function at the valleys are −8 and −6.

5/5 Question 9 : 5 ptsFor the graph below of f(x), f ' (0) is

For the graph below of f(x), f ' (0) is

positive negative zero undefined

The graph is decreasing at x = 0 so f ' is negative there.

5/5 Question 10 : 5 ptsSkip to question text.

Find the vertical asymptotes.y=x2+2xx3−4xFind the vertical asymptotes.y=x2+2xx3−4x x=2 x = −2, x = 0 x = −2, x = 0, x = 2 There are no vertical asymptotes.

Start by factoring and canceling.

5/5 x 2 + x (x + 2x 2) y = Question 11 : 5 ptsSkip to question text. = x 3 − x (x 2 − Find the vertical asymptotes of 4x 4) f(x).f(x)=x2−x−2x2+x−2Find the vertical asymptotes of f(x).f(x)=x2−x−2x2+x−2 x (x + 2) 1 x = −1, x = 2 , x ≠ 0, x ≠ x = −1 = = x (x − 2) (x − − 2. x = 1, x = −2 (x + 2) 2) f (x) has no vertical asymptotes.

To find vertical asymptotes, start byfactoring both the numerator anddenominato Now set the denominator equal to zero. r to see if any factorscan be canceled.f(x)=x2−x−2x2+x−2=(x−2)(x+1)(x+2) x−2 (x−1)No factors cancel, so find those pointswhere the denominator equals zero. (x = 0 +2)(x−1)=0x=1 and x=−2So, the lines x=1 and x=– x= 2 arevertical asymptotes.To find vertical asymptotes, start byfactoring both the nume 2. rator anddenominator to see if any factorscan be canceled.f(x)=x2−x−2x2+x−2=(x−2 Verify that the numerator is not )(x+1)(x+2) equal to (x−1)No factors cancel, so find those pointswhere the denominator equals zero. (x+2) zero at this point (it isn’t) and (x−1)=0x=1 and x=−2So, the lines x=1 and x=–2 arevertical asymptotes. you’re done. 5 / 5 Question 12 : 5 ptsFind the absolute maximum and absolute minimum values of f (x) = x 3 − 6x 2 + 9x − 3 on the interval [−1, 2].

Find the absolute maximum and absolute minimum values of f (x) = x 3 − 6x 2 + 9x − 3 on the interval [−1, 2].

Absolute maximum value: 1 Absolute minimum value: −3 Absolute maximum value: none Absolute minimum value: −3 Absolute maximum value: 1 Absolute minimum value: −19 Absolute maximum value: none Absolute minimum value: −19

None of these

Use the Closed Interval Method. First find the critical points: f ′(x) = 3x 2 − 12x + 9 3x 2 − 12x + 9 = 0 when x = 3 or x = 1. x = 3 is not in the interval [−1, 2], so x = 1 is the only critical point to be considered.

Next, evaluate the function at the critical points and the endpoints. f (1) = 13 − 6 (1)2 + 9 (1) − 3 = 1 f (−1) = (−1)3 − 6 (−1)2 + 9 (−1) − 3 = −19 f (2) = 23 − 6 (2)2 + 9 (2) − 3 = −1 The largest value, 1, is the absolute maximum value. The smallest value, −19, is the absolute minimum value.

5/5 Question 13 : 5 ptsSkip to question text.

What are the critical points of the function f(x)=x+1x?What are the critical points of the function f(x)=x+1x? x = −1, 0, 1 x = −1, 1 x=1 x=0

The derivative is f′(x)=1−1x2.To determine the critical points, find all of the points where the derivativeequals zero, a nd all of the pointswhere the derivative does not exist,but the original function does.Set f ′(x)=0 to find the critical poi ntsx=−1, 1. The derivative does notexist at x=0, but neither does the original function. The only critical pointsare x =−1, 1.The derivative is f′(x)=1−1x2.To determine the critical points, find all of the points where the derivativeequals zer o, and all of the pointswhere the derivative does not exist,but the original function does.Set f′(x)=0 to find the critical poin tsx=−1, 1. The derivative does notexist at x=0, but neither does the original function. The only critical pointsare x=−1, 1 .

5/5 Question 14 : 5 ptsSkip to question text.

Given f (1) = −3 and f (5) = 9. If f is continuous on [1, 5] and differentiable on (1, 5), the mean value theorem guarantees which of the following statements about f for some c in the interval (1, 5). f (c) = 0 f ′(c) = 0 f (c) = 3 f ′(c) = 3

Since f is continuous on [1, 5] and differentiable on (1, 5), then the mean value theorem guarantees there is at least one c in the interval (1, 5) such that f ′(c) equals the average rate of change over the interval. That is, for some c in the interval (1, 5), f ′(c) = 3. f′(c)=f(5)−f(1)5−1=9−(−3)4=3f′(c)=f(5)−f(1)5−1=9−(−3)4=3 5/5

Question 15 : 5 ptsFind the absolute maximum and absolute minimum values of f (x) = 4x −1 on the interval [−2, 1].

Find the absolute maximum and absolute minimum values of f (x) = 4x −1 on the interval [−2, 1].

Absolute maximum value: 4 Absolute minimum value: −2 Absolute maximum value: 4 Absolute minimum value: none Absolute maximum value: none Absolute minimum value: −2 Absolute maximum value: none Absolute minimum value: none None of these

The Closed Interval Method cannot be used because the function is not continuous on the closed interval.

As x approaches 0 from the right, the function approaches positive infinity.

As x approaches 0 from the left, the function approaches negative infinity.

There is no absolute maximum value and no absolute minimum value.

5/5 Question 16 : 5 ptsSkip to question text.

Which of the curves has the following characteristics? f ′(−2.8) = 0 f ″(−2) = 0 f ″(−0.25) = 0 for x < −2, f ″(x) > 0 for x > −0.25, f ″(x) > 0

This curve has minima or maxima ( f ′(x) = 0) at x = −2.8, −0.5 and 0. Its second derivative is 0 at x = −2 and x = −0.25 ( f ″ = 0) and these are the locations of inflection points. At values of x less than −2 and greater than −0.25, the second derivative is positive and the curve is concave up ( f ″(x) > 0).

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

Find the interval(s) where the functionf(x)=1x2 is increasing.Find the interval(s) where the functionf(x)=1x2 is increasing. x≥0

x>0 x < 0 or x > 0 x<0

Step 1: Find f ′ (x).

1 f ( x) . x = 2

− f ′(x) = −2x 2 −3

=

.

x3 Step 2: Find the critical points of f (x) by finding those values of x where f ′(x) is either zero or undefined. − f ′(x) 2 . = x3 f ′(x) will never equal zero but it is undefined at x = 0. Note: Technically, x = 0 is not a critical point since it is not in the domain of the function. But we should still include it in our analysis! You’ll see why in this example.

Step 3: Make a sign chart by finding the value of f ′(x) for a point in each

regio n: 5/5 − f ′(x) 2 . = Question 18 : 5 ptsSkip to question text. x3 Below is the sign chart for a complicated function that has 6 critical points. How many maxima, − minima, and critical pionts that are neither does this function have? f ′(1) 2 = −2, so f ′(1) = < 0.

3 1 2 minima, 3 maxima, 1 neither −2 − 3 minima, 2 maxima, 1 neither 2 f ′(−1) = 2, so f ′(−1) 0 minima, 1 maximum, 5 neither = (−1 = > 0.

3 minima, 3 maxima, 0 neither − 3 Minima occur wherever the derivative changes from negative to positive (going left to right), and this ) 1 occurs in three places, at critical points x = 1, x = 3, and x = 6. Thus, there are three minima. Maxima occur Step 4: f ′(x) > 0 when x < 0, where the derivative changes from positive to negative (going left to right), and this occurs at x = 2 and x = so that 5. Thus, there are two maxima. At x = 4, the derivative does not change signs, so that critical point is neither is the interval where f (x) is a maximum nor a minimum. increasing.

Note the importance of the 5 / 5 point x = 0!

Question 19 : 5 ptsSkip to question text.

Which of the following curves is the graph of the equation f(x)=−x2+x+6x2+x−2?f(x)=−x2+x+6x2+x−2?

Since the function is a rational fuction and can be simplified, start by finding the domain restrictions and cancelling. f (x) =

−x 2 + x + 6

x2 + x − 2 −(x − 3) (x + 2)

= (x − 1) (x + 2) x− 3 = ,x≠ − −2 x− 1

Now you are ready to find the asymptotes. (x − 1) = 0 → x = 1; vert. asymptote x− x− 3 3 li li = = − − m m x → x − −1, x → − x − −1 ∞ ∞ 1 1 → y = −1; horiz. asymptote

Now take the derivative to see where the function is increasing or decreasing.

Notice that t2h e derivative is always negative, so the function must be continuously decreasing. Plotting a few points will show that graph C is the best ,x≠ cf h′ o(ixc)e.= − (x − − 2 5 /5 2 1)

Question 20 : 5 ptsThe graph of f(x) is given below. f(x) is increasing on the approximate interval(s):

The graph of f(x) is given below. f(x) is increasing on the approximate interval(s):

(-infinity,-4) (-4,3.2) (3.2,infinity) (-infinity,-4) and (3.2,infinity)

5/5

Quiz Score: 100 out of 100