Module 2 Mastery Exercises - 1

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Apr 3, 2024

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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 1/32 Module 2 Mastery Exercises QUESTION 1 · 1/1 POINTS The graph of function is shown below. At which value of is the slope of the tangent line to the curve equal to ? BACK TO OVERVIEW Attempts Attempt 1: 67% (10/15 points), Feb 06 at 9:55pm CST Questions to show: All (15)
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 2/32 That is correct! Answer Explanation Correct answer:
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 3/32 Checking the slope of the tangent at each point, we see that the slope of the tangent at is approximately equal to . So the answer is . QUESTION 2 · 1/1 POINTS Given the function , which of the following is a valid formula for the instantaneous rate of change at ? That is correct! FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 4/32 Answer Explanation Correct answer: Recall that the instantaneous rate of change of a function at is given by Therefore, the instantaneous rate of change of at is given by QUESTION 3 · 1/1 POINTS Find the slope of the secant line between and on the graph of the function . That is correct! FEEDBACK Content attribution −2
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 5/32 Answer Explanation This question is asking for the rate of change, which is the same as the slope between two points on the curve. When , the -value on the curve is So the ±rst point is . When , the -value on the curve is So the second point is . The average rate of change is the slope between these points, The average rate of change between and for the function is . In other words, the secant line that connects these points has slope . QUESTION 4 · 1/1 POINTS Find the values of and that make the following piecewise function di²erentiable everywhere. Correct answers: −2 FEEDBACK Content attribution
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 6/32 That is correct! Answer Explanation Correct answer: and Note that both of the pieces of this piecewise function are polynomials, which are always di²erentiable. The only point we are unsure about is the boundary point . So we need to make sure the function is di²erentiable at . Before we can worry about di²erentiability, we must make sure the function is continuous. The limits from the left and right must be equal for the function to be continuous. The limit of the function from the left is The limit of the function from the right is Setting these limits equal and solving for , we ±nd and and and and and and
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 7/32 As long as this condition is satis±ed, we will have that the limit exists and . Now, for the function to be di²erentiable, the limit for the derivative must exist. Because the function is piecewise, we must consider the limit from the left and from the right and make sure that they are equal. The limit of the di²erence quotient from the left is The limit of the di²erence quotient from the right is Setting the derivatives from the left and the right equal to each other, we ±nd So we have found that . Plugging this into the equation we found earlier for , we ±nd
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 8/32 So the ±nal answer is and . QUESTION 5 · 1/1 POINTS The graph of is shown below. Find the graph of . That is correct! FEEDBACK Content attribution
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 9/32
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 10/32 Answer Explanation Correct answer:
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 11/32 Notice that to the left of , the graph is a line with slope . Thus, the derivative to the left is the constant function . To the right of , the graph is a parabola which opens upward. So the derivative to the right is a line with positive slope. Note that the derivative of is not de±ned at because the graph has a corner there, so there is an open circle at . The graph of and its derivative are shown together below.
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 12/32 QUESTION 6 · 1/1 POINTS Given the graph of below, ±nd the graph of the derivative of . FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 13/32 That is correct!
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 14/32
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 15/32 Answer Explanation Correct answer:
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 16/32 We can see that the graph has turning points at and , where the tangent is horizontal, and so the derivative is . The graph has negative slope (and the derivative is negative) to the left of and to the right of , and the slope is positive (and the derivative is positive) between and . Putting this together, we ±nd the graph is a parabola with zeroes at and which opens downward.
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 17/32 QUESTION 7 · 1/1 POINTS Find from using the de±nition of a derivative. That is correct! Answer Explanation Correct answer: FEEDBACK Content attribution
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 18/32 The de±nition of the derivative is: Substitute and into the de±nition: Use the common denominator to combine the fractions in the numerator: Expand the numerator: Simplify: Cancel out the common factor, : Evaluate the limit: This can be simpli±ed:
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 19/32 Therefore, QUESTION 8 · 1/1 POINTS Using the de±nition of a derivative, ±nd given . That is correct! Answer Explanation Use the de±nition of a derivative: Substitute and : FEEDBACK Content attribution s ′ x = −2 x − 8 ( ) Correct answers: s ′ x = −2 x − 8 ( )
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 20/32 Expand the binomials: As we combine the expanded polynomials, all of the terms without have opposites, so they cancel. Further expand and simplify: Factor out : Cancel out the common factor : Evaluating the limit gives . QUESTION 9 · 0/1 POINTS Estimate the instantaneous rate of change of the function at using the average rate of change over successively smaller intervals. That's not right. FEEDBACK Content attribution
2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 21/32 Answer Explanation Correct answer: We will compute the average value over the intervals , , , and . Computing over the interval , we ±nd Now, over the interval , we ±nd Next, over the interval , we ±nd Finally, over the interval , we ±nd
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 22/32 So we can see that the average rate of change over these successively smaller intervals appears to be approaching . Your answer: QUESTION 10 · 0/1 POINTS Given the graph of below, what is the open interval(s) over which is positive? Give your answer in interval notation. For example, . FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 23/32 That's not right. Answer Explanation We can see that the graph has turning points at and . The graph has negative slope (and the derivative is negative) to the left of and to the right of , and the slope is positive (and the derivative is positive) between and . Therefore, the answer is . QUESTION 11 · 1/1 POINTS Why is the function shown below not di²erentiable at ? −2, 3 Correct answers: −2,3 ( ) FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 24/32 That is correct! Answer Explanation Correct answer: The function is not continuous at the point. The function has a vertical tangent line at the point. The function has a horizontal tangent line at the point. The function has a corner at the point. The function is smooth at the point. The function is not continuous at the point. The function is continuous at the point.
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 25/32 Remember that a function cannot be di²erentiable if it is not continuous. In this case, there is a hole at , so that is the reason the function is not di²erentiable at . QUESTION 12 · 0/1 POINTS Consider the graph of shown below in blue. Submit your answer to this question by completing the following tasks: 1. Drag the movable black point to a point where the derivative of the function is . 2. Draw an approximation for the tangent line to the function at your chosen location by dragging the red point. That's not right. FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 26/32 Answer Explanation Checking the slope of the tangent line at each point, we see that the slope of the tangent at is approximately equal to (this result turns out to be exact for the shown curve). Therefore, . QUESTION 13 · 1/1 POINTS Find from using the de±nition of a derivative. That is correct! FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 27/32 Answer Explanation The de±nition of the derivative is: Substitute and into the de±nition: Multiply the numerator and the denominator by without distributing the denominator: Simplify the numerator: q ′ ( x ) = 2 7 x − 2 7 √ Correct answers: q ′ ( x ) = 2 7 x − 2 7 √
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 28/32 Cancel out : Evaluate the limit: QUESTION 14 · 0/1 POINTS At which of the following -values does the function shown below have negative slope? Select all correct answers. FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 29/32 That's not right. Answer Explanation Correct answer: If we draw tangents at the di²erent points, we can see that the slopes are negative at the points and .
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 30/32 Your answer: QUESTION 15 · 0/1 POINTS Suppose is continuous at , and that . FEEDBACK Content attribution
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 31/32 Which of the following must be true? Select two correct answers. That's not right. Answer Explanation Correct answer: . exists. Remember that if a function is di²erentiable at the point , then it must also be continuous at . However, the converse is not true. In other words, it is possible for a function to be continuous at a point but not di²erentiable . To help you remember this, think about examples like the absolute value function, which is continuous at its "corner" but not di²erentiable. Therefore, all we know about is that the limit at exists and equals the function value: Your answer: is di²erentiable at . . exists. . exists. .
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2/6/24, 10:07 PM Module 2 Mastery Exercises - Knewton https://www.knewton.com/learn/section/ef980133-667b-4ae2-a540-118a4c17192f/quiz/79317b97-3997-4586-94a7-4aec778e7fc7/results 32/32 . FEEDBACK Content attribution
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