MAT 141 Solutions – Exam 4 (Applications of Differentiation)

1. To prove the inequality, analyze the graph of the function �(�) = .

() From the derivative function �′(�) = = , we have �(�) = 0 () () whenever � + 4� − 2 = 0, or � = −� ± √�. These are the only critical numbers of � since (2 + �) > 0 for all real values of �.

Now, �(�) > 0 on −2 − √6, −2 + √6 and �(�) < 0 on −∞, −2 − √6 and on (−2 + √6, ∞). Applying the first derivative test, we know that � attains a local minimum at � = −2 − √6 equal to �−2 − √6 = √ ≈ −0.11 > −0.12 and � attains a local maximum at � = −2 + √6 equal to �−2 + √6 = √ ≈ 1.11 < 1.2.

Since � is a proper rational function, its graph (shown below) approaches the �-axis at both ends (i.e. � → 0 as � → ±∞). This, in conjunction with the previous findings, implies the inequality 2 + � −0.12 ≤ ≤ 1.2 ∎ 2 + �

Note: The end behavior of � also implies that both local extrema are, in fact, absolute extrema of the function.

2. Let � = � − � tan(10�) = �(�), then �′(�) = 1 − . Since � is a function that () is continuous and differentiable for all real values of � and � has local extrema at � = ±0.3, Fermat’s Theorem implies the following:

�(±0.3) = 0 ⇒ 1 − = 0 ⇒ = 1 ⇒ � = �. (±)

Below is the graph of �(�) = � − 10 tan � with its two absolute extrema at � = ±0.3.

3. First, we compute the following:

1 1 �(5) − �(0) − 1 = 4 9 = . 5 5 36

Since �′(�) = , solving the equation �′(�) = yields ()

−2 1 = ⇒ (� − 3) = −72 ⇒ � = 3 + √−72 ≈ −1.16 (� − 3) 36

Since 3 + √−72 < 0, there is no value � in the interval (0, 5) such that �′(�) = . This result is still consistent with the Mean Value Theorem since the function � is discontinuous at � = �. 4. a) First-Derivative Test

Applying the Quotient Rule, we compute the derivative function of � as follows:

4�� − �� 4� − � �(4 − �) �(�) = = = . (�) � �

Hence �(�) = 0 for � = 0 or � = 4. These are the only two critical numbers of � since � > 0 for all real values of �.

Since �(�) < 0 for � < 0, �(�) > 0 for 0 < � < 4, and �(�) < 0 for � > 4, the function is decreasing on the intervals (−∞, 0), (4, ∞), and increasing on the interval (0, 4). This implies that � has a local minimum at � = 0 equal to �(0) = 0 and a local maximum at � = 4 equal to �(4) = ≈ 4.689.

b) Test for Concavity

Applying the Quotient Rule again to �′, we compute the second derivative function of � as follows:

(12� − 4�)� − (4� − �)� �(�) = (�)

12� − 8� + � �(� − 2)(� − 6) = = . � �

Hence �(�) = 0 for � = 0, � = 2, or � = 6. These are the only critical numbers of �′ since � > 0 for all real values of �.

Since �(�) > 0 for � < 0, 0 < � < 2, and � > 6, the graph of � is concave up on the intervals (−∞, 0), (0, 2) and (6, ∞). Since �(�) < 0 for 2 < � < 6, the graph of � is concave down on the interval (2,6). As a result, the graph of f �� ���� has 2 inflection points at �, ≈ (�, �. ���) and �, ≈ (�, �. ���). �� ��

[You could summarize all the findings in parts a) and b) in sign tables for the derivative and second derivative, respectively.]

c) Left end behavior: lim �(�) = lim �� = ∞ → → (“∞ × ∞” determinate form)

Right end behavior: lim � (�) = lim = lim = lim = lim = � → → → → →

(by repeated use of L’Hôpital’s Rule with "∞/∞" indeterminate form)

d) Below is the graph of � showing all extrema, inflection points, and end behavior.

5. First, note that �(�) = ln(1 − 2� + �) = ln(1 − �) = 2 ln(1 − �)

� () () � � Then, � (�) = = and so � (�) = = = −� �. () () ���

Since > 0 for all � ≠ ±1, we conclude that the graph of � (shown below) has no () inflection point.

6. Let �(�) = cos � − sin �. We then seek to approximate the root of the equation �(�) = 0 on the interval [0, �] using Newton’s method.

Since �′(�) = −sin � − sin 2�, we use the recursive formula

cos � − sin � � = � + (� = 0, 1, 2, 3, … . ) sin � + sin 2�

starting with � = 1.

The computations yield the following approximations:

� ≈ 0.904173 � ≈ 0.904557 � ≈ 0.904557

We thus conclude that the solution of the equation cos � = sin �, for 0 ≤ � ≤ �, is approximately �. ����.

7. a) Since (ln �) → ∞ as � → ∞, we recognize the indeterminate form “∞/∞”.

We then apply L’Hôpital’s Rule as follows:

1 (ln �) 2(ln �) � 2 ln � lim = lim = lim . → � → 1 → �

Once again, we apply L’Hôpital’s Rule since we have the same indeterminate form:

2 ln � 2 lim = lim = � → � → �

b) Since lim (csc � − cot �) = lim − = lim , we recognize the ⟶ ⟶ ⟶ indeterminate form “0/0”. We then apply L’Hôpital’s Rule as follows:

lim = lim = = � ⟶ ⟶

c. First, note that () () lim lim (1 − 2�) = lim � = �→ . → →

We recognize the indeterminate form “0/0” with this limit. By L’Hôpital’s Rule, we then have −2 ln(1 − 2�) 2 lim = lim 1 − 2� = lim = −2 → � → 1 → 2� − 1

We thus conclude that

� lim (1 − 2�) = � = → ��

8. Let � be the line that passes through (3, 5) and cuts off the least area from quadrant I.

Then the equation of � is given in slope-point form by � − 5 = �(� − 3), where � is the slope of �.

Since � crosses the �-axis at � = 5 − 3� and � crosses the �-axis at � = (check this), the triangular area we seek to minimize is given by

1 3� − 5 −(3� − 5 ) −��� + ��� − �� �(�) = (5 − 3� ) = = , 2 � 2� ��

where � < 0.

Differentiating this area function, we get

(30 − 18�)(2�) − 2(−9� + 30� − 25) 50 − 18� �(�) = = 4� 4�

� Since 4� > 0 for � < 0, we have �(�) = 0 whenever 18� = 50, or � = − . �

We therefore conclude that the equation of � is given by

� � − 5 = − (� − 3) ⟹ � = − � + �� �

Bonus Questions

B1. Let �(�) = � + � + � + 1. Then �(−1) = −2 and �(0) = 1. By the Intermediate Value Theorem, the equation �(�) = 0 has a real root � on the interval (−1, 0).

The derivative of this polynomial function is given by

�(�) = 101� + 51� + 1 = 101� + 51� + 1,

where � = �.

Since �(�) = 0 yields only negative solutions for � (check this), we conclude that there are no real values of � that solve the equation �(�) = 0 . this implies that the function � has no local extrema on ℝ. Thus, there is no possibility of the equation �(�) = 0 having another real root anywhere (if you assume, on the contrary, that there is another root somewhere, say at � = �, then �(�) = �(�) = 0 and this contradicts Rolle’s Theorem). ∎

B2. To prove that � > �, it suffices to prove that � > e ∙ ln � (apply the natural base logarithm to both sides of the inequality), or � − e ∙ ln � > 0.

Let �(�) = � − e ∙ ln �. We then need to show that �(�) > 0.

Since �(�) = 1 − , we have �(�) = 0 when � = �. Check now that �(�) = 0 is an absolute minimum of �. It then follows that �(�) > 0. ∎

B3. The first and second derivatives of � = are given by

1 − 2� − � �(� − �)(�� + �� + �) � = and � = (1 + �) (� + ��)�

Solving �′′ = 0 and applying the test for concavity yields the following three inflection ponts:

� − √� � + √� (�, �), −� − √�, , −� + √�, � �

� � Check that these three IPs lie on the line � = � + as shown in the graph below. � �