1. society
2. crime
3. property crime
4. usury

Math 226
Exam #2 Sample Problems
Sections 3.1-3.6, 4.1, 4.2, 4.4
1. The graph below represents a function f (x). State whether the sign of f , f 0 , and f 00 at
each of the points marked on the graph is positive, negative, or zero.
A
B
C
D
f
+
+
+
+
f0
0
+
0
f 00
+
0
-
2. Find all extrema of the function on the given interval and state what the absolute
extrema are.
a) f (x) = 10 + 27x − x3 on [0, 4]
To find the extrema, we need to find critical points and test them along with the end
points.
f 0 (x) = 27 − 3x2 = 3(9 − x2 ) = 3(3 + x)(3 − x)
Setting f 0 (x) = 0 and solving for x, we see that x = ±3. So we have two stationary points
and these are our only two critical points, because our derivative exists everywhere.
Since −3 is not on [0, 4], we only need to test the critical point x = 3, along with the end
points 0 and 4.
If we look at how the sign of the derivative changes around the end points and the
stationary point, we see that x = 3 is a local, or relative maximum, and because the function
is increasing to the right of x = 0 and decreasing to the left of x = 4, they are both local
(relative) minima.
So it seems as if x = 3 is going to be our global, or absolute maximum, but we should
still double check.
f (0) = 10, f (3) = 64, f (4) = 54
This tells us that x = 0 yields the absolute (global) minimum, and x = 3 yields the
absolute maximum.
b) g(x) =
f 0 (x) =
x2
x
on [−2, 0]
+x+1
1(x2 + x + 1) − (2x + 1)(x)
(x2 + x + 1)2
=
−x2 + 1
(x2 + x + 1)2
The denominator is never going to be zero. This means that the only critical points we
might have will be stationary points.
So if we set f 0 (x) = 0 and solve for x, then we end up only solving for −x2 + 1 = 0, and
we have x = ±1. However since 1 is not part of our interval, we only need the stationary
point x = −1.
This means we have to look at f (−1), f (−2) and f (0) for our extrema. First let’s use the
first derivative test to find local max/min.
The sign chart shows that x = −1 would yield a local minimum, and that both x = −2
and x = 0 would yield local maxima.
Now to find the absolute extrema, note that f (0) = 0, f (−1) = −1, and f (−2) = − 23
This means that x = −1 yields an absolute minimum, and x = 0 yields an absolute
maximum.
π π
c) h(x) = x − tan x on − ,
4 4
π π
h(x) exists everywhere on the interval − , , so we only need to look for stationary
4 4
points.
h0 (x) = 1 − sec2 x = 1 −
1
cos2 x
.
1
h0 (x) = 0 whenever 1 =
⇒ cos2 x = 1 ⇒ cos x = ±1.
cos2 x
π π
Over the interval − , , cos x = 1 at x = 0 and cos x , −1.
4 4
So x = 0 is the only point of concern other than the endpoints. However, h is decreasing
on both sides of x = 0, so this point is also no longer a potential extremum. We are left
with the endpoints.
Since h is decreasing on the right of −π/4 and decreasing to the left of π/4, we see that
x = −π/4 is the maximum and x = π/4 is the minimum. Of course I’ll double check...
π
π
π
π
π
π
π
π
h −
= − − tan −
= − + 1 and h
= − tan
= −1
4
Thus x = −
4
π
4
4
4
4
is the absolute maximum and x =
4
π
4
4
4
is the absolute minimum.
3. Find an equation for the line tangent to the curve given by 5x2 − 6xy + 5 y2 = 16 at the
point (1, −1).
We use the for of the tangent line equation y = f (x0 ) + f 0 (x0 )(x − x0 ), where (x0 , f (x0 )) =
(1, −1)
So we are going to have to take the derivative, but we must do so implicitly as
follows:
5x2 − 6xy + 5 y2
= 16
!
dy
dy
+ 10 y
10x − 6 1 · y + x ·
dx
dx
= 0
dy
dy
+ 10 y
dx
dx
= 0
10x − 6 y − 6x
dy
(−6x + 10 y) = 6 y − 10x
dx
dy
dx
At the point (1, −1), we have
=
6 y − 10x
10 y − 6x
dy 6(−1) − 10(1)
=
=1
dx (1,−1) 10(−1) − 6(1)
This tells us that f 0 (x0 ) = 1. So, our equation is
y = −1 + 1(x − 1) = −1 + (x − 1)
4. A 14 ft. ladder is leaning against a wall. The top of the ladder is being lifted along the
wall away from the floor at a constant rate of 3 ft/s.
a) How fast will the bottom of the ladder be moving toward the wall when the bottom
of the ladder is 5 ft away from the wall?
Note the following picture.
From the picture, we can see that x, h, and 14 are related by the pythagorean theorem.
So we have
x2 + h2 = 142
2x
dx
dh
+ 2h
dt
dt
= 0
Now we know that x = 5, because we want dx/dt when the bottom of the ladder is 5 ft
away from the wall. Using the pythagorean theorem, we have
√
√
52 + h2 = 142 ⇒ h = 142 − 52 = 3 19 ≈ 13.08
So we have
2(5)
√
dx
+ 2(3 19)(3) = 0
dt
dx
dt
=
√
−18 19
10
≈ −7.85 f t/s
b) How fast is the angle between the top of the ladder and the wall changing when the
bottom of the ladder is 5 ft away from the wall.
dθ
when x = 5 (using x from our original picture). θ is adjacent to both
dt
h and the ladder, so we can use the relationship
We are finding
cos θ =
So, −14 sin θ ·
sin θ =
5
14
h
14
or 14 cos θ = h
dh
dθ
= 1 . Note that when x = 5 and the hypotenuse is 14, we have that
dt
dt
.
Therefore, we have −14
5
14
dθ
dθ
3
= 1 (3), or
= − = −0.6 rad/s.
dt
dt
5
5. If a snowball melts so that its surface area decreases at a constant rate of 1 cm2 /min,
find the rate at which the diameter decreases when the diameter is 10 cm.
2
D
2
The surface area of a sphere (snowball) is AS = 4πr = 4π
= πD2
2
So if we know that
dAS
dD
= −1, we want to find
when D = 10. So,
dt
dt
AS = πD2
dAS
dt
= π(2D)
dD
dt
−1 = π(2)(10)
−
1
20π
cm/min =
dD
dt
dD
dt
6. For each of the following, find: i) intervals on which f is increasing and decreasing,
ii) all critical points, iii) local and absolute extrema, iv) Inflection points and intervals of
concavity.
We need to find where the derivative is positive/negatve, where it equals zero/DNE, use
the first or second derivative test, find where the second derivative is zero and test points
around it.
a) f (x) = 2 − 2x − x3
i. f 0 (x) = −2 − 3x2 < 0 for all x, telling us that f is decreasing on (−∞, ∞).
ii. Since f 0 (x) < 0 for all x, f (x) has no critical points.
iii. Without any critical points, f has no local extrema, and without a closed interval, it
will also have no absolute extrema.
iv. f 00 (x) = −6x. The only possible inflection point is where f 00 (x) = 0, which is when
x = 0. We consider a sign chart for f 00 ::
Since f 00 changes sign as we go through x = 0, this says that x = 0 is in fact an
inflection point, and this also tells us that f is concave up on (−∞, 0) and concave down
on (0, ∞).
b) f (x) = x4 + 4x3
i. When f 0 (x) = 0, we have x = 0, −3. So we look at the behavior of f 0 to see where f is
increasing/decreasing.
The sign chart for f 0 shows that f is increasing on [−3, ∞) and decreasing on
(−∞, −3].
ii. Since the derivative exists everywhere, our only critical points are at x = 0, −3.
iii. Using our first derivative sign chart we see that x = −3 is a local minimum and x = 0
is not an extremum. Also, since the graph is decreasing to the left of −3 and increasing
to the right, f (−3) gives us the smallest value on (−∞, ∞) so it is also a global minimum.
There is no absolute maximum.
iv. f 00 (x) = 12x2 + 24x = 12x(x + 2)
When f 00 (x) = 0, we have x = 0, −2. So these are our possible inflection points. We
consider the sign of f 00 .
Looking at the sign chart for f 00 , we see that both x = −2 and x = 0 are inflection points
for f , and f is concave up on (−∞, −2] ∪ [0, ∞) and concave down on [−2, 0].
c) f (x) =
i. f 0 (x) =
1
1 − x2
0 − (−2x)(1)
(1 − x2 )2
=
2x
(1 − x2 )2
f 0 (x) = 0 when x = 0, and f 0 (x) DNE when x = ±1. We make a sign chart for f 0 .
The sign chart for f 0 shows us that f is increasing on [0, ∞) and decreasing on
(−∞, 0].
ii. The critical points are at x = 0, ±1.
iii. Using the sign chart for f 0 , we can conclude that there is a local minimum at x = 0,
and no local maximum. Since x = 0 yields the smallest point, we can also conclude that
x = 0 gives our absolute minimum, and we have no absolute maximum.
d) f (x) = cos2 x − 2 sin x on [0, 2π]
i. f 0 (x) = 2 cos x(− sin x) − 2 cos x = 2 cos x (− sin x − 1)
π 3π
On our given interval, f 0 (x) = 0 when cos x = 0, or x = ,
, and also when
2 2
3π
− sin x − 1 = 0 ⇒ sin x = −1 or x =
.
2
Hence f is increasing on
π 3π
π
3π
,
and decreasing on 0,
∪
, 2π
2
2
2
ii. f 0 (x) exists everywhere, so our only critical points are at x =
2
π 3π
,
2
2
π
iii. The first derivative test (see sign chart for f 0(x)) tells us that x = is a local minimum
2
3π
is a local maximum.
and x =
2
π
π
π
= cos2
− 2 sin = −2, and
Note also that f (0) = f (2π) = cos2 (0) − 2 sin 0 = 1, f
2
2
2
3π
3π
2 3π
=2
f
= cos
− 2 sin
2
2
2
3π
The absolute maximum happens at x =
and the absolute minimum happens at
2
π
x= .
2
iv. f 00 (x) = −2 sin x (− sin x − 1) + 2 cos x (− cos x) = 2 sin2 x + 2 sin x − 2 cos2 x =
2 sin (x) + 2 sin(x) − 2(1 − sin2 (x)) = 4 sin2 (x) + 2 sin(x) − 2 = 2(2 sin(x) − 1)(sin(x) + 1)
2
Note that 2 sin(x) − 1 = 0 at x =
π 5π
3π
,
and sin(x) + 1 = 0 at x =
.
6
6
2
Sign chart attack!
π
Sew, f is concave up on π6 , 56π and concave down on 0, π6 ∪ 56π , 2π with both x =
6
5π
and
as inflection points.
6
e) f (x) =
1
2
ex
2
= e−x
2
i. f 0 (x) = −2xe−x . This is zero at x = 0. If we test points around 0, we find that f 0 (x) is
positive to the left of 0, and negative to the right of 0.
Soh f is increasing when x ∈ (−∞, 0] and decreasing when x ∈ [0, ∞)
ii. The derivative exists everywhere. The only critical point is at x = 0.
iii. Using the second derivative test, we have
2
2
2
f 00 (x) = −2 e−x − 2x2 e−x = −2e−x 1 − 2x2
f 00 (0) = −2(1)(1) = − telling us that x = 0 gives a local maximum. In part (i) we found
that f was increasing to the left of x = 0 and decreasing to the right of x = 0, we can
conclude that x = 0 would also yield an absolute maximum, and we have no absolute or
local minima.
1
iv. f 00 = 0 when 1 − 2x2 = 0, so when x = ± √ .
2
1
1
1
1
f 00 > 0 when x < − √ and x > √ and f 00 < 0 when − √ < x < √ .
2
2
2
1
2
1
f has two inflection points, at x = ± √ , is concave up on −∞, − √
2
2
!
1
1
concave down on − √ , √
2
f) f (x) = tan−1 x2
i. f 0 (x) =
!
∪ √ , ∞ and
1
2
2
1
1+
!
(x2 )2
(2x) =
2x
1 + x4
f 0 (x) = 0 when x = 0. If we test points around our stationary point, we find that f 0 (x)
is positive to the right of zero and negative to the left. Thus f is increasing on [0, ∞) and
decreasing on (−∞, 0].
ii. f 0 (x) exists everywhere. Our only critical point is at x = 0.
iii. From part (i) we can conclude that x = 0 gives a local and absolute minimum. There
is no absolute maximum due to the way that the function behaves in terms of increasing
and decreasing around x = 0.
iv. f 00 (x) =
2(1 + x4 ) − 4x3 (2x)
(1 + x4 )2
=
2 − 6x4
(1 + x4 )2
=
2(1 − 3x4 )
(1 + x4 )2
1
This is zero when x = ± √
4
3
Testing points around these values tells us that f is concave up on − √41 ,
3
1
concave down on −∞, − √41 ∪ √41 , ∞ , with two inflection points at x = ± √
4
3
3
3
1
√
4
3
and
g) f (x) = x2 ln(x)
i. f 0 (x) = 2x ln x + x = x (2 ln x + 1). This is zero when x = 0, e−1/2 . However, since x is
only differentiable for values greater than zero, x = 0 is eliminated.
h
f 0 (x) < 0 on (0, e−1/2 ) and f 0 (x) > 0 on (e−1/2 , ∞), so f is increasing on e−1/2 , ∞ and
i
decreasing on 0, e−1/2 .
ii. x = e−1/2 is the only critical point in the domain of f .
iii. x = e−1/2 is both a local and absolute minimum. There is no absolute or local
maxima. The point at x = 0 might look like a local minimum, but on an open interval, 0
is not an option.
2
iv. f 00 (x) = (2 ln x + 1) + x
= 2 ln x + 3
x
00
This is zero when
x = e−3/2 . Using a sign chart
for f (x) would tell us that f is concave
−
3/2
−
3/2
up on e
, ∞ and concave down on 0, e
.
7. What is an equation of the line tangent to the curve y = x3 − 3x2 + 4x at the curve’s
inflection point?
To find the inflection point we should take the second derivative, and test points. Then
we can put the tangent line equation together.
dy
d2 y
= 3x2 − 6x + 4 and 2 = 6x − 6 = 6(x − 1)
dx
dx
This is zero when x = 1. The second derivative also changes from negative to positive
as it goes through x = 1, verifying that x = 1 is in fact an inflection point.
Using our original function, we find that when x = 1, y = 13 − 3(1)2 + 4(1) = 2
dy Also,
= 3(1)2 − 6(1) + 4 = 1
dx x=1
Therefore, the equation of the tangent line at the inflection point, (1, 2), is
y = 2 + 1 (x − 1)
8. At what point(s) is the line tangent to the curve y2 = 2x3 perpendicular to the line
4x − 3 y + 1 = 0?
If we solve for y in 4x − 3 y + 1 = 0, we have y =
line is
4
3
1
3
4
+ x, telling us that the slope of this
3
, and the slope of a line perpendicular to this one would be −
This tells us that we need
dy
dx
3
4
= − 43 . So we need to find dy/dx.
y2 = 2x3
2y
dy
dx
= 6x2
y
dy
dx
= 3x2
dy
dx
=
−
3
4
=
3x2
y
3x2
y
y = −4x2
Now we can substitute this value of y into the equation y2 = 2x3 and solve for x:
y2 = 2x3
2
−4x2
= 2x3
16x4
8x4
8x4 − x3
= 2x3
= x3
= 0
x (8x − 1) = 0
3
x = 0,
1
8
Now substituting these values into the equation y = −4x2 , we see that y = 0 and
dy
1
y = − . But
does not exist at x = y = 0.
16
dx
Therefore the point at which
the tangent line to the curve y2 = 2x3 is perpendicular to
1
the line 4x − 3 y + 1 = 0 is
8
,−
1
16
.
9. Find the local linearization of the function at x0 .
For these we are using the equation f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 )
a) f (x) = x3 at x0 = 1
Here, we have f (x) ≈ x30 + 3x20 (x − x0 ). So using x0 = 1 yields
f (x) ≈ 13 + 3(1)2 (x − 1) = 1 + 3(x − 1)
b) g(x) = cos x at x0 =
π
2
π
Here, g(x) ≈ cos x0 + (− sin x0 ) (x − x0 ). For x0 = we have:
2
π
π
π
π
g(x) ≈ cos
− sin
x−
=− x−
2
c) k(x) =
√
2
2
2
1 + x at x0 = 0
√
This time, k(x) ≈
1 + x0 + √ 1
2
√
k(x) ≈
1+x0
1+0+
(x − x0 ). When x0 = 0, we have
√
2
1
1+0
(x − 0) = 1 +
1
2
(x − 0)
10. Each problem below displays the graph of a function. Roughly sketch the function’s
derivative. Pay attention to the scaling of the axes!
a) The derivative is the cubic...the quartic is the original.
b) The derivative is in bold. In the derivative, there is a vertical asymptote at x = 2.
11. A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes
directly over a radar station. Find the rate at which the distance from the plane to the
station is increasing when it is 2 mi away from the station.
Using the picture above, we can relate x and y by the Pythagorean theorem as
follows:
x2 + 12 = y2
2x
dx
dt
= 2y
dy
dt
When y = 2, we have that
x2 + 12 = 22 ⇒ x =
√
4−1=
√
3
So,
√
2(
3)(500)
√
250
3
= 2(2)
=
dy
dt
dy
dt
√
Thus dy/dt = 250 3 ≈ 433 mi/h.
12. Suppose water is being poured into an upside-down conical container in which the
radius r is always half the height, h. If the height of the water level is increasing at a rate
of 5 ft/min, at what rate is the water pouring into the container when the water level is 10
ft. high.
We are finding
V=
1
πr2 h,
3
dV
when h = 10 ft.
dt
however since r =
1
2
h, we can rewrite V =
dV 3h2 dh h2 dh
=
=
dt
12 dt
4 dt
When h = 10, we have
dV 102
=
· 5 = 125 ft3 /min.
dt
4
1 h
3 2
!2
h=
h3
12
13. Find
dy
dx
a) x3 − y3 = 6xy
x3 − y3 = 6xy
dy
3x − 3 y
dx
2
2
3x2 − 3 y2
dy
dx
3x2 − 6 y
dy
= 6 1·y+x
dx
= 6 y + 6x
= 3 y2
3x2 − 6 y
3 y2
+ 6x
x2 − 2 y
y2 + 2x
dy
dx
dy
dy
+ 6x
dx
dx
=
dy
dx
=
dy
dx
x
b) xe = y
xe
x
= y
e ln x = ln y
x
ex ln x + ex
1
!
=
x
ex
y ex ln x +
=
x
x
ex
xe ex ln x +
=
x
1 dy
y dx
dy
dx
dy
dx
c) y = (ln x)2
y = (ln x)2
d) y =
p
r
3
= 2 (ln x)1
dy
dx
=
1
x
2 ln x
x
cos−1 (x2 )
y =
e) y =
dy
dx
p
cos−1 (x2 )

−1/2 
− p
cos−1 x2

dy
dx
=
dy
dx
= −p
1
2
1
1 − (x2 )


 (2x)
2
x
(cos−1 (x2 )) (1 − x4 )
x2 − 1
x2 + 1
r
y =
ln y =
1 dy
y dx
dy
dx
dy
dx
=
3
x2 − 1
x2 + 1
1
3
1
ln x2 − 1 − ln x2 + 1
1
3 x2 − 1
= y·
1
1
3 x2 − 1
r
=
(2x) −
1 

3
3
1
x2 + 1
(2x) −
(2x)
1
x2 + 1
(2x)

x2 − 1 
1
1
(
(
2
x)
−
2
x)

x2 + 1  x2 − 1
x2 + 1
f) 2x3 y − y2 = 3
g) y = 2xe
2x3 y − y2
= 3
!
dy
dy
− 2y
2 3x · y + x
dx
dx
= 0
2
3
6x2 y +
dy
(2x3 − 2 y) = 0
dx
dy
dx
=
−6x2 y
2x3 − 2 y
dx
dy
=
−3x2 y
x3 − y
√
x
√
y = 2xe
−1
h) y = esin
x
dy
dx
√ 1
√
= 2 1 · e x + x · e x x−1/2
dy
dx
= 2e
2
√
x
+
√ √x
xe
x
dy
−1
1
= esin x √
dx
1 − x2
!
i) y = ln tan−1 (3x)
!
dy
1
1
3
=
(3) =
−
1
2
2
dx tan (3x) 1 + (3x)
(1 + 9x ) (tan−1 (3x))
14. Find the limits. If you use L’Hopital’s rule, make sure you note this.
ln x
x−1
The numerator and denominator have limits of zero. So, we can apply L’Hopital’s
rule.
a) lim
x→1
1
L0 H
=
lim
x→1
=
lim
x→1
=
b) lim−
x→π
sin x
1 − cos x
=
0
2
x
1
1
x
1
= 0. Note that L’Hopital’s rule does not apply.
ln x
c) lim √
x→∞ 3 x
The numerator and denominator have limits of ∞. So, we can apply L’Hopital’s
rule.
L0 H
=
=
=
=
1
lim
x→∞
lim
x→∞
lim
x
1
3x2/3
3x2/3
x
3
x→∞ x1/3
0
d) lim− sec x − tan x
x→ π2
sec x − tan x
=
=
lim
x→(π/2)−
(sec x − tan x)
=
1
cos x
=
sin x
cos x
1 − sin x
cos x
lim
x→(π/2)−
L0 H
=
−
lim
x→(π/2)−
1 − sin x
cos x
− cos x
− sin x
0
L’Hopital’s rule applied because the numerator and denominator both have limits of
0.
ex − 1 − x
x→0
x2
L’Hopital’s rule applies (twice actually).
e) lim
L0 H
=
L0 H
=
=
ex − 1
x→0 2x
lim
lim
x→0
1
2
ex
2
15. Below is the graph of y = f 0 (x) on the interval [−1, 3.5].
a) Estimate the intervals on which f is increasing. Justify your answer.
f 0 > 0 on (0, 1) and (3, 3.5), so f is increasing on [0, 1] ∪ [3, 3.5].
b) Estimate the locations of all inflection points for f in the interval [−1, 3.5]. Explain
your answer.
An inflection point, where the second derivative changes signs, would happen at
a maximum or minimum of the first derivative...where the first derivative changes from
increasing to decreasing, the second derivative would change from positive to negative
and go through zero. The same happens for the minimum on the derivative.
So the inflection points are at x ≈ 0.45, 2.2
16. The graph below includes a function f , its derivative f 0 , and its second derivative f 00 .
Identify which is which and explain your answers.
A
B
C
Explanation: The easiest thing to check is where the stationary points on one function are
zeros on another...this indicates a function and its derivative.
Looking at A, its second stationary point could be a zero on C, but the first stationary
point does not have a zero for either of the others. So C cannot be the graph of the
derivative of the function whose graph is A.
Looking at B, its first and second stationary points are zeros on C, so C can be the graph
of the derivative of the function whose graph is B.
For C, its stationary points are zeros on A, So A can be the graph of the derivative of
the function whose graph is B.
Therefore A is f 00 , B is f , and C is f 0 .
17. Use an appropriate local linear approximation to estimate the value of the given
quantity.
√
3
a) 26.8
Here we will use f (x) =
√
3
x = x1/3 , and x0 = 27. So,
f (x) ≈
near x0 = 27: f (x) ≈
√
3
√
3
x0 +
27 +
= 3+
f (26.8) ≈ 3 +
1
27
1
27
1
2/3
3x0
(x − x0 )
1
3 (27)
2/3
(x − 27)
(x − 27)
(26.8 − 27)
≈ 2.993
b) (5.1)2
This time we will use f (x) = x2 and x0 = 5. Sew,
f (x) ≈ x20 + 2x0 (x − x0 )
near x0 = 5: f (x) ≈ (5)2 + 2(5)(x − 5)
= 25 + 10(x − 5)
f (5.1) ≈ 25 + 10(5.1 − 5)
= 26
c) cos (0.1)
Now, f (x) = cos x, and x0 = 0. Sough,
f (x) ≈ cos x0 − sin x0 (x − x0 )
near x0 = 0: f (x) ≈ cos 0 − sin 0 (x − 0)
= 1