Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
1
Problem 1. Let f ( x ) = 2 x3 − 8 .
(a) Use the limit of difference quotients to find the instantaneous rate of change of the function
at the point x = 3 .
(b) Find an equation of the tangent line to the graph of f at x = 3 .
(c) Is f increasing or decreasing at that point? Explain.
Solution.
2(3 + ∆x)3 − 8 ) − ( 2 ⋅ 33 − 8 )
(
f (3 + ∆x) − f (3)
= lim
(a) f ′(3) = lim
∆x → 0
∆x → 0
∆x
∆x
3
2
2 ⋅ 3 + 6 ⋅ 3 ∆x + 6 ⋅ 3( ∆x) 2 + 2( ∆x)3 − 8 − 2 ⋅ 33 + 8
= lim
∆x → 0
∆x
2
2
6 ⋅ 3 ∆x + 6 ⋅ 3( ∆x) + 2( ∆x)3
= lim
∆x → 0
∆x
2
= lim 6 ⋅ 3 + 6 ⋅ 3∆x + 2( ∆x) 2
∆x → 0
= 6 ⋅ 32 = 54
(b) f (3) = 2 ⋅ 33 − 8 = 46 so we want the equation of a line through the point (3,46) with slope 54.
That is y = 46 + 54( x − 3) .
(c) Since the derivative f ′(3) is positive, f has a positive slope at x = 3 , thus f is increasing at
this point.
Problem 2. For the function f whose graph y = f ( x ) is given below, sketch the graphs of the
derivative y = f ′( x ) and the antiderivative* y = F ( x ) , satisfying an additional condition that
F (0) = 2 .
*Recall this means F ′( x ) = f ( x ) .
Solution. To sketch the graph of y = f ( x ) we first notice that for large negative values of x the
graph of f has a very small negative slope, and so its derivative is negative, but very close to
zero. As x is increasing, the slope of f first becomes more negative, then it is zero at the local
minimum at x = 2 (and so the graph of f ′ has an x-intercept at that point). Then f is increasing
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
2
until x = 0 and decreasing again until x = 2 , this gives us two more x-intercepts for the graph of
f ′ . At x = 3 the graph of f has a corner – the slope jumps from a positive slope of
approximately 1 to a constant slope of zero, so f is not differentiable at x = 3 and the graph of
f ′ will have a jump discontinuity at that point:
y = f ( x)
Slopes here
become
y = f ′( x )
Heights here
To sketch the graph of the antiderivative y = F ( x ) we first notice that f is zero at approximately
x = ±1 . Since the values of f change from negative to positive at these points, the function F
will have a local minimum at x = −1 . Similarly, it will have a local maximum above x = 1 . This
is also in synch with the fact that at x = 0 the slope of F is positive. Taking into account that
F (0) = 2 we can begin sketching our graph. Note that for x > 3 the slope of F becomes constant
and is equal to −1 , and for large negative values of x the slope of F is almost zero (since the
values of f are almost zero), and so the graph of F levels off:
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
3
y = F ( x)
Slopes here
y = f ( x ) = F ′( x )
Heights here become
Problem 3. An orange is thrown into the air. Its height at various times is shown in the table.
t (seconds)
s(t) (feet)
0
3
0.5
26.5
1
42.0
1.5
49.5
2
49.0
2.5
40.5
3
24.0
3.5
0
(a) Find the average velocity of the orange between t = 2 and t = 3 seconds.
(b) Use the table below to find the best possible underestimate for the instantaneous velocity of
the orange exactly one second after it is released.
t (seconds)
s(t) (feet)
0
3
0.9
0.99
0.999
39.54 41.756 41.975
1.0
42
1.001
1.01
1.1
42.023 42.228 44.14
(c) Explain how your results above (both of them) can be visualized on a graph of s (t ) .
2
49
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
4
Solution.
(a) Average velocity =
∆s 24 ft − 49 ft
=
= −25 ft s
∆t
1s
(b) From the values in the table, we see that the orange is still going up (s is increasing) at t = 1
s. Thus the orange will be slowing down at this time and average speeds after t = 1 will give
underestimates for the instantaneous speed at t = 1 . The best approximation will come from
choosing as small a time interval as possible:
ds ∆s 42.023 ft − 42 ft
≈
=
= 23 ft s
dt ∆t
0.001 s
(c) Both results above are slopes of secant lines on the graph of s vs. t. Specifically the slope of
the line between the points (2,49) and (3,24) on the graph is –25. The slope of the line
between the points (1,42) and (1.001,42.023) on the graph is 23. The instantaneous speed at
t = 1 corresponds to the slope of the tangent line at the point (1,42) on the graph.
Problem 4. Use the definition of the derivative (as a limit) to find a formula for the derivative of
g ( x ) = 1 + 1 / x . Math grammar counts! (Reporting the derivative without using the definition to
arrive at your answer will earn you no credit.)
Solution.
(1 + x +∆1 x ) − (1 + 1x )
g ( x + ∆x) − g ( x)
= lim
∆x → 0
∆x → 0
∆x
∆x
1
1
−
= lim x +∆x x
∆x → 0
∆x
g ′( x) = lim
= lim
∆x → 0
= lim
∆x → 0
x −( x +∆x )
( x +∆x ) x
∆x
−∆x
( x +∆x ) x
∆x
= lim
−1
∆x → 0 ( x +∆x ) x
=
−1
x2
Problem 5. Suppose C ( r ) is the total cost of paying off a car loan borrowed at an annual interest
rate of r %.
(a) What are the units of C ′( r ) ?
(b) What is the sign of C ′( r ) ? Explain briefly.
(c) What is the practical meaning of C ′(5.0) = 1352.77 ? Your statement should talk about
paying off a car loan.
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
5
Solution.
(a) dollars per percent
(b) C ′( r ) > 0 because increasing the annual interest rate, r, will increase the cost of paying off
the loan, C ( r ) . That is, C ( r ) is an increasing function.
(c) C ′(5.0) = 1352.77 is the approximate additional cost, in dollars, of paying off a loan if the
interest rate on the loan is increased from 5% to 6%. Thus the 1% increase in the interest rate
will result in an additional cost of approximately $1352.77 to repay the loan.
 x sin ( 1x ) , x ≠ 0
Problem 6. Graph f ( x) = 
on your calculator. Use the result to decide if f is
x=0
 0,
differentiable at x = 0 . Explain your answer.
Solution. The graph looks like the following, even after continually zooming in. Thus it never
looks like a straight line with a defined slope. This function is not differentiable.
 x 2 sin ( 1x ) , x ≠ 0
Problem 7. Graph g ( x) = 
on your calculator. Use the result to decide if f is
x=0
 0,
differentiable at x = 0 . Explain your answer.
Solution. The graph looks like the following at two different zoom levels. We can see both
graphically and algebraically that the function is bound between y = − x 2 and y = x 2 , causing it
to flatten out horizontally at x = 0 . Continuing to zoom in bears this out. So g is differentiable at
x =0.
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
Problem 8. Approximate the slope of the tangent line to the graph of y = x x at x = 2 to within
0.1. Show your work.
Solution. From the graph below, we can clearly see that the function is increasing and concave
up at x = 2 (“concave up” means the slope is increasing). This means that the slopes of secant
lines taken slightly to the left of x = 2 will generate underestimates to the slope of the tangent
line at x = 2 . Likewise that the slopes of secant lines taken slightly to the right of x = 2 will
generate overestimates to the slope of the tangent line at x = 2 .
Thus
slope ≈
1.91.9 − 2 2
= 6.1443 is an underestimate,
1.9 − 2
and
2.12.1 − 22
slope ≈
= 7.4964 is an overestimate.
2.1 − 2
Unfortunately, this only allows us to bound the
error by the difference of the two, or 1.358.
Thus we are not within 0.1 of the actual slope.
Refining our approximations by taking smaller
changes, we try
slope ≈
1.991.99 − 22
= 6.7057 is an underestimate, and
1.99 − 2
6
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
7
2.012.01 − 22
slope ≈
= 6.8404 is an overestimate.
2.01 − 2
Unfortunately, our underestimate and overestimate still differ by more than 0.1, so we cannot be
sure that these are within the desired error bound.
Refining yet again, we compute
slope ≈
1.9951.995 − 22
= 6.7390 is an underestimate, and
1.995 − 2
slope ≈
2.0052.005 − 2 2
= 6.8064 is an overestimate.
2.005 − 2
Now the difference between the underestimate and overestimate is 0.067 which is less than the
desired error bound. So either 6.7390 or 6.8064 will serve as an approximation accurate to within
0.1 of the slope of the tangent line at x = 2 .
Note: We will discuss techniques later in the course that would allow us to compute that the
slope of the tangent is exactly 4(1 + ln 2) .
Problem 9. State the limit definition of the derivative of a function f.
Solution. Any of the following forms of the definition would suffice as an answer:
f ′( x ) = lim
f ( x + ∆x ) − f ( x )
∆x
f ′( x ) = lim
f ( x + h) − f ( x )
h
∆x → 0
h→0
f ′( x0 ) = lim
x → x0
f ( x) − f ( x0 )
x − x0
Problem 10. Fill in blanks with the appropriate
expressions from the definition of the derivative to
label the quantities marked on the graph of
y = f ( x ) as illustrated below.
y
y = f (x)
A
E
D
F
G
H
x
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
8
Solution.
A = f ( x + ∆x )
B = f ′( x ) or lim
∆x → 0
C=
f ( x + ∆x ) − f ( x )
∆x
f ( x + ∆x ) − f ( x )
∆x
D = f ( x)
E = f ( x + ∆x ) − f ( x )
F = x + ∆x − x or ∆x
G=x
H = x + ∆x
Problem 11. Suppose g (v) is the fuel efficiency, in miles per gallon, of a car going at v miles
per hour.
(a) What are the units of g ′(45) ?
(b) What is the practical meaning of the statement g ′(45) = −0.62 ?
(c) Given that g (45) = 27 and g ′(45) = −0.62 , estimate g (48) and explain the practical
meaning of your answer.
Solution.
(a) Since g (v) is measured in miles per gallon (mpg) and v is measured in miles per hour (mph),
g ′(45) is measured in mpg/mph.
(b) The statement g ′(45) = −0.62 means that the fuel economy at this velocity is a decreasing
function of v, and increasing the velocity by 1 mph results in the decrease of the fuel
efficiency by approximately 0.62 mpg.
(c) Increasing the speed from 45 mph to 48 mph is a change of 3 mph. At the given rate of
−0.62 mpg/mph, this means the fuel efficiency would change by approximately
−0.62 ⋅ 3 = −1.86 mpg. Thus the fuel efficiency when traveling at a speed of 48 mph will be
g (48) ≈ 27 − 1.86 = 25.14 mpg.
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
9
Problem 12. Suppose that f is a function with f (125) = 40 , f ′(125) = 6 , and f ′′( x ) > 0 for all
x.
(a) Estimate f (126.5) .
(b) Is your answer to Part (a) an underestimate or overestimate. Explain.
Solution.
(a) The statements f (125) = 40 and f ′(125) = 6 tell us that at x = 125 , the value of the function
is 40 and the function is increasing at a rate of 6 units for a unit increase in x. Since we
increase x by 1.5 units in going from 125 to 126.5, the value of the function goes up by
approximately 1.5 ⋅ 6 = 9 units, so f (126.5) ≈ 40 + 1.5 ⋅ 6 = 40 + 9 = 49 .
(b) The statement that f ′′( x ) > 0 for all x tells us that the graph of f is concave up (the slopes are
increasing). Thus the tangent line at x = 125 will lie under the graph of f. Consequently using
the tangent line to approximate values approximate values of f ( x ) will produce
underestimates.
Problem 13. Suppose I am playing with a toy car and the total distance the car is away from me
is given by the following function, = ln + 1 + √ + 1, where is measured in feet and is given in seconds. Observe that the car will slow down as time goes on, which means that the
graph will be increasing and concave down. Without using any derivative rules use the
following table to approximate the instantaneous rate of change at = 2:
t (secs)
s (ft)
1.9
1.99
2
2.01
2.1
2.76764939359
2.82443503398
2.83066309624
2.83687523605
2.89208379766
(Warning: Do not round!)
(a) Describe what you are trying to approximate in the context of the problem and assign it an
appropriate derivative symbol.
(b) Find an overestimate and an underestimate for the instantaneous rate of change at = 2.
(c) Write an algebraic expression for the error for both the overestimate and underestimate.
(d) Find an error bound using your overestimate and underestimate you found in part b).
(e) Find an approximation with error less than 0.001 and explain how you know.
Solution.
(a) I am trying to approximate the instantaneous velocity of the car at 2 seconds. Velocity is
change in distance divided by the change in time. Using an appropriate derivative symbol,
we can call this unknown ′2.
(b) We can compute approximations to the instantaneous rate of change by using average rates
of change. Using ∆t = ±0.01 , we get
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
10
2.83066309624 2.82443503398
= 0.622806226
2 1.99
2.83687523605 2.83066309624
= 0.621213981
2.01 2
We know that the first estimate is an over estimate because the graph for the given function
is always concave down (the slopes are decreasing) so that slopes to the left will be smaller.
Similarly we can see that the second estimate is an under estimate because the graph of the
function is always concave down so that slopes to the right will be larger.
(c) error = |′2 0.622805226|
error
!"
= |′2 0.621213981|
(d) errorbound = |0.622805226 0.621213981| = 0.001591245
(e) If we use t=2.00001 we can see that we get an approximation with error less than 0.01
because of we consider the following approximations we see
2.83066309624 2.0001 2.83066309624 2.83072529629
=
= 0.6220005
0.0001
0.0001
and
1.9999 2.83066309624 2.83060089459 2.83066309624
=
= 0.6220165
0.0001
0.0001
We can see that that the error bound associated with these over and underestimate is
0.000016 and that any choice of t between 2.0001 and 1.9999 will have an error less than
0.000016, which will result in an error less than 0.001. (NOTE: The number I pick was a
little extreme, but it works.)
Problem 14. Sketch the graph of a continuous function that has all of the following properties:
• & ' ( ) 0for 0 * ( * 4
• & '' ( ) 0 for 0 * ( + 4
• & ' ( ) 0 for 4 + ( * 10
• & '' ( + 0 for 4 + ( * 10
• &( has an x-intercept at ( = 4.
Solution.
Math 131 Fall 2013
CHAPTER 2 EXAM (PRACTICE PROBLEMS - SOLUTIONS)
11
Problem 15. A continuous function & , defined for all (, has all of the following properties:
• &is decreasing
• &is concave up
• &7 = 4
• & ' 7 = ,
-
Sketch a possible graph for & , and use it to answer the following questions about & .
(a) &1 + 0,
&1 = 0 ,
,
(b) & ' 1 + ,
-
&1 ) 0,
,
& ' 1 = ,
-
or “we can’t tell”
,
& ' 1 ) ,
-
(circle one)
or “we can’t tell”
(circle one)
(c) lim0→23 &( =_______
Solution.
(a) &1 ) 0. Since the graph of f is concave up, it lies above the tangent line at ( = 7 and
must cross the x-axis somewhere to the right of where the tangent line does. Since the
,
slope of the tangent line at ( = 7is , it crosses the x-axis at ( = 1. Thus f must be
positive at ( = 1.
,
(b) & ' 1 + . Since &isconcaveupso the slope of & is increasing as x increases, or
-
,
alternatively the slope is decreasing as x decreases, therefore & ' 1 + .
-
(c) lim0→23 &( = ∞. Since f is decreasing, it will have larger and larger values as we
look to the left. Since &is concave up, the changes in &( also increase (for equal
changes in x). Thus the graph continues to climb more and more steeply as we move to
the left. So, lim0→23 &( = ∞.