4.1
Concept: Extreme Values
Question 1:
Worked out Solution:
Question 2:
Worked out solution:
Concept: Critical Values
Question 3:
4.2
In questions 1
a). state whether or not the function satisfies the hypotheses of the Mean Value Theorem on the given
interval
b). if it does, find each value of c in the interval (a,b) that satisfies the equation.
1. f(x)=ln(x-1) on [2,4]
Answer:
In question 2 a≤x≤b is given.
Let A=(a,f(a)) and B=(b,f(b)).
Write an equation for
•
The secant line AB
•
A tangent line to f in the interval (a,b) that is parallel to AB.
2. f(x)=√(x-1) 1≤x≤3
Answer:
Concept: Find the Anitderivative
3. y'=1/(x-1) at (2,5)
y=ln(x-1)+c
5=ln(2-1)+c
5=c
y=ln(x-1)+5
4.3 (For additional help, go to pages 205-214 in the book)
Concept: 1st Derivative Test
Question:
Use the First Derivative Test to determine the local extreme values of the function, and identify any
absolute extrema.
y={
3-x2 , x<0
x2 +1, x≥0
Solution:
Concept: Concavity
Question:
Use the Concavity Test to determine the intervals on which the graph of the function is (a) concave up
and (b) concave down.
y=4x3+21x2+36x-20
Solution:
Concept: Concavity
Question:
Use the Concavity Test to determine the intervals on which the graph of the function is (a)
concave up and (b) concave down.
y=5-x1/3
Solution:
Concept: Point(s) of Inflection
Question:
Use the derivative of the function y=f(x) to find the points at which f has a point of inflection.
y’=(x-1)2(x-2)
Solution: Just look at letter (c) for the answer since we have formatted the question a bit
differently than the book!
4.4
Concept: Solve geometric application problems
Question:
Answer:
Concept: Solve physical application problems.
Question:
Solution:
Concept: Solve trigonometric application problems.
Question:
Answer:
4.5
Find the linearization L(x) at x=a
4. f(x)=√𝑥 2 + 9 a=2
Answer
Approximate the root by using a linearization centered at an appropriate nearby number.
3
5. √26
3
f(x)= √𝑥
x=27
Tangent Line at (27,3) is f '(27)=1/3(27)-2/3
f '(27)=1/3 x 1/9= 1/27
Answer
In question 6 use Newton's Method to estimate all real solutions of the equation. Make sure your
answer accurate to 6 decimal places
6. x4+x-3=0
Answer
4.6 (For additional help, go to pages 246-250 in your book)
Concept: Hypotenuse
Question:
A balloon is rising vertically above a level straight road at a constant rate of 1 ft/sec. Just when
the balloon is 65 ft above the ground, a bicycle moving at a constand rate of 17 ft/sec passes
under it. How fast is the distance between the bicycle and balloon increasing 3 sec later?
Concept: Angle Theta
Question:
A and B are walking on straight streets that meet at right angles. A approaches the intersection
at 2 m/sec and B moves away from the intersection at 1 m/sec. At what rate is the angle 
changing when A is 10 m from the intersection and B is 20 m from the intersection? Express
your answer in degrees per second to the nearest degree.
Solution:
Concept: Cones
Question:
Sand falls from a conveyor belt at the rate of 10 m3/min onto the top of a conical pile. The
height of the pile is always three-eighths of the base diameter. How fast are the (a) height and
(b) radius changing when the pile is 4 m high? Give your answer in cm/min.
Solution:
Unit 2 Review
Find the average rate of change of the function over the given interval
7. f(x)=x2+2x at [1,7]
f(x)=(1)2+2(1)
f(x)=3
f(x)=(7)2+2(7)
f(x)=63
(1,3) (7,63)
(63-3)/(7-1)=60/6=10
Unit 3 Review
Question: Find the derivative algebraically.
1a. 8x7+9x6-3x3-42
1b. 2+𝟐𝒙
1c. 𝟐𝒙 + 3
a. 0
b. 56x6+54x5-9x2
c. 56x6+54x5-9x2-8
d. 15x6+15x5
a. DNE
b. -2
c. -2x
−1
d. 2𝑥 2
a. 2x
b. 0
c. 2𝑥 ln2
d. 2𝑥 ln3
2a. sin(4x2+5x)
2b. tan-1(3x2)
2c. cos2(7x)
a. cos(4x2+5x)
b. sin(8x+5)
c. (8x+5)sin(4x2+5x)
d. –sin(4x2+5x)
𝟏
a.
6𝑥
1 + 9𝑥 4
3x2
b. 1+9x4
1
c. 1−9𝑥 4
d.
1
1+9𝑥 4
a. –sin2(7x)
b. -7sin(7x)cos(7x)
c. cos(7)
d. -14sin(7x)cos(7x)
3a. ln(𝟐𝒙 )
1
3b. 𝒍𝒐𝒈𝟐 𝟓𝒙𝟑
15𝑥 2
a. 2𝑥
a. 5𝑥 3 𝑙𝑛2
b. 2
c. ln2
b. 5𝑥 3 𝑙𝑛2
1
a. 2e
b. 0
c. e2
d. 2ee’
1
5𝑥 3
c. 𝑙𝑛2
2𝑥
d. 𝑙𝑛2
3c. e2
1
d. 𝑙𝑛2
Solutions:
1a. b. 56x6+54x5-9x2
Use the Product Rule.
Don’t be fooled that the derivative of 42 is 8 due to the Product Rule.
42 is actually equal to 16 and the derivative of a constant is always 0.
−1
1b. d. 2𝑥 2
Think of the problem like this instead: 2+(2x)-1 so you can easily use the power rule.
Using the power rule, you should get –(2x)-2(2). The extra 2 at the end comes into place by the
chain rule because the derivative of 2x is 2. Simplify this all and you should get:
−𝟏
𝟐𝒙𝟐
1c. 𝟐𝒙 + 3
c. 2𝑥 ln2
This is an exponential function. The derivative of (𝒂𝒖 ) is 𝒂𝒖 ln 𝒂 (u)’.
Plug this expression into the formula with a=2 and u=x, you should get 𝟐𝒙 ln𝟐 (x)’ which
equals 𝟐𝒙 ln𝟐.
2a. c. (8x+5)sin(4x2+5x)
The derivative of sin u is cos u times (u)’. In this case (4x2+5x)=u. So the derivative of
sin(4x2+5x)=cos(4x2+5x) times (4x2+5x)’ which equals (8x+5)sin(4x2+5x).
6𝑥
2b. a. 1+9𝑥 4
𝟏
𝟏
The derivative of arctan u is 𝟏+𝒖𝟐 (𝒖)′. Substitute u for 3𝒙𝟐 and you should get 𝟏+(𝟑𝒙𝟐 )𝟐 (𝟑𝒙𝟐 )′
𝟔𝒙
which equals, in its most simplified form, 𝟏+𝟗𝒙𝟒 .
2c.d. -14sin(7x)cos(7x)
Rewrite this problem as [cos(7x)][cos(7x)]. Use the product rule now. It should look like this:
[cos(7x)]’[cos(7x)]+[cos(7x)][cos(7x)]’.
This now equals, in its most simplified form using trigonometric derivatives:
-7sin(7x)cos(7x)-7sin(7x)cos(7x)
=-14sin(7x)cos(7x)
3a. c. ln2
𝟏
𝟏
The derivative of ln u equals 𝒖 (𝒖)′. Now just substitute u for 𝟐𝒙 . You should get 𝟐𝒙 (𝟐𝒙 )′ . The
derivative of 𝟐𝒙 is used by the derivative of exponential functions. The derivative of (𝒂𝒖 ) is
𝒂𝒖 ln 𝒂 (u)’.
Plug this expression into the formula with a=2 and u=x, you should get 𝟐𝒙 ln𝟐 (x)’ which
𝟏
𝟏
equals 𝟐𝒙 ln𝟐. Now you can put it all together. 𝟐𝒙 (𝟐𝒙 )′ =𝟐𝒙 𝟐𝒙 𝒍𝒏𝟐=
𝟐𝒙 𝒍𝒏𝟐
𝟐𝒙
=ln2.
15𝑥 2
3b. a5𝑥 3 𝑙𝑛2
1
The derivative of 𝑙𝑜𝑔𝑎 𝑢 = 𝑢 ln 𝑎 (𝑢)′. Now substitute 5x3 for u and 2 for a, and 15x2 for (u)’ since
1
that is the derivative of 5x3. When it is put all together, you should get (5𝑥 3 )(ln 2) (15𝑥 2 ) which
15𝑥 2
equals 5𝑥 3 𝑙𝑛2.
3c. b. 0
Do not use the power rule! Remember, e is a CONSTANT! All derivatives of constants are ZERO!
Don’t make the same mistake as the 𝜋15 quiz!