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Math 105: Review for Exam II - Solutions
- Find dy/dxdy/dxdy/dx for each of the following.
(a) y = x
2
x
2
2 x
- ln 2 + ln (2x) + arctan 2
dy
dx
= 2x + (ln 2)
x
2 x
2 x
· 2 Note that e
2
, ln 2, and arctan 2 are constants.
(b) y =
x · arctan (5x)
dy
dx
x
− 1 / 2
arctan(5x) +
x ·
1 + (5x)
2
arctan(5x)
2 x
1 / 2
x
1 + 25x
2
(c) y = ln(tan(
cos(x
2
)
dy
dx
tan(
cos(x
2
)
· sec
2
cos(x
2
)
) · ln 2(
cos(x
2
)
) · (− sin(x
2
)) · 2 x
(d) y = sin
5
x + e
π
ln 4 + arcsin 6x
Note that sin
5
w = (sin w)
5
dy
dx
= 5 sin
4
x + e
π
ln 4 + arcsin 6x
·cos
x + e
π
ln 4 + arcsin 6x
(1)(ln 4 + arcsin 6x) − (x + e
π
1
1 −(6x)
2
(ln 4 + arcsin 6x)
2
- Consider the curve defined by x
3
3
x xy
3
3
x xy
3
3
xy (known as the Folium of Descartes).
(a) Find dy/dxdy/dxdy/dx. Use implicit differentiation.
3 x
2
2
dy
dx
y +
x
dy
dx
3 y
2
dy
dx
x
dy
dx
y − 3 x
2
dy
dx
3 y
2
x
y − 3 x
2
dy
dx
9
2
y − 3 x
2
3 y
2
9
2
x
(b) Verify that the point (1,2) is on the curve above.
We must check to see if the values x = 1 and y = 2 satisfy the equation above.
x
3
3
?
xy
3
3
?
?
Thus, the point (1,2) is on the curve.
(c) Find the equation of the tangent line at the point (1,2).
We want y = mx + b.
m =
9
2
2
2
9
2
, so y =
x + b.
Now plug in x = 1 and y = 2 to find b.
· 1 + b ⇒
= b
Therefore, we have y =
x +
- Evaluate the following limits.
Throughout this solution, the symbol F will stand for whatever notation your instructor prefers for
using L’Hopital’s Rule on the indeterminate form 0/0; this may be
“0/0”
= or
L
′
H
= or
H
= or = “0/0” or
“has the form ‘
’ and so, by L’Hopital’s Rule, is equal to” or something else. The symbol ♥ will serve
the same purpose for the indeterminate forms ∞/∞ and −∞/∞.
(a) lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
lim
x→ 1
x
3
7 − 7 x
F lim
x→ 1
3 x
2
(b) lim
x→ 0
1 − cos 2x
x
lim
x→ 0
1 − cos 2x
x
lim
x→ 0
1 − cos 2x
x
= 0 Can’t use (and don’t need) L’Hopital’s Rule!
(c) lim
x→ 0
x
2
ln x
lim
x→ 0
x
2
ln x lim
x→ 0
x
2
ln x
This is of the indeterminate form 0 · (−∞) so we rewrite the function as a fraction in order to use
L’Hopital’s Rule.
lim
x→ 0
x
2
ln x = lim
x→ 0
ln x
1
x
2
♥ lim
x→ 0
1
x
− 2
x
3
= lim
x→ 0
x
x
3
= lim
x→ 0
x
2
(d) lim
x→ 0
1 − cos 4x
5 x
2
lim
x→ 0
1 − cos 4x
5 x
2
lim
x→ 0
1 − cos 4x
5 x
2
F lim
x→ 0
4 sin 4x
10 x
F lim
x→ 0
16 cos 4x
(e) lim
x→∞
x
2
x
lim
x→∞
x
2
x
lim
x→∞
x
2
x
♥ lim
x→∞
2 x
ln 2 · 2
x
♥ lim
x→∞
ln 2 · ln 2 · 2
x
- Suppose the domain of fff(((xxx))) is all reals and that fff has an inverse function f
− 1
f (x)
− 1
f (x)
− 1
(x). Further,
suppose that fff(2) = 5(2) = 5(2) = 5 and f
′
f(2) = e
′
f(2) = e
′
(2) = e. Finally, let hhh(((xxx) = 1) = 1) = 1/f/f/f(((xxx))).
(a) What point must be on the graph of f
− 1
f (x)
− 1
f (x)
− 1
(x)?
Since f(2) = 5, we also know that f
− 1
(5) = 2; therefore, the point (5,2) is on the graph of f
− 1
(x).
(b) What point must be on the graph of h(x)
h(x) h(x)?
Since f(2) = 5, we also know that h(2) = 1/5; therefore, the point (2,1/5) is on the graph of h(x).
The moral of the story in parts (a) and (b) is that inverses and reciprocals are not the same.
(c) Give an example of a point that cannot be on the graph of f(x)
f(x) f(x). Do not choose a
point with x
x x-value of 2.
Since f
− 1
(x) is a function, we know that for each of its inputs (such as 5) there is only one output
(in this case, 2). This means that no point of the form (5, k) can be on the graph of f
− 1
(x)
(except for k = 2). Therefore, no point of the form (k, 5) can be on the graph of f(x) (except
for k = 2). [Note: this is the same as saying that f must be a one-to-one function; on f, no two
x-values correspond to the same y-value.]
(d) What is the value of the derivative of h(x)
h(x) h(x) at x = 2
x = 2 x = 2?
By the Quotient Rule (or Chain Rule if you prefer), h
′
(x) =
0 · f(x) − 1 · f
′
(x)
[f(x)]
2
f
′
(x)
[f(x)]
2
So, h
′
f
′
[f(2)]
2
e
2
e
- Suppose that y = f(t)
y = f(t) y = f(t) is a solution to the differential equation y
′
π
arcsin t + y
2
y
′
π
arcsin t + y
2
y
′
π
arcsin t + y
2
and that
f
f
f
. Find the equation of the tangent line to fff at
We want y = mx + b.
m =
π
arcsin
2
π
π
, so y =
x + b.
(d) Find the x
x x-value(s) of all inflection points.
f
′′
(x) = 12x
2
e
x
3
e
x
3
e
x
4
e
x
Use Product Rule on each product in f
′
(x) above.
0 = e
x
(x
4
3
2
0 = e
x
x
2
(x
2
0 = e
x
x
2
(x + 2)(x + 6)
⇒ x = 0, − 2 , − 6
x < − 6 − 6 < x < − 2 − 2 < x < 0 0 < x
f
′′
positive negative positive positive
f concave up concave down concave up concave up
So, the x-values of the inflection points of f are x = −2 and x = −6 but NOT x = 0.
(e) Sketch f
f f.
0
-12 -6 -4 -2 0
4
10
8
6
2
- How would your answers to the previous question change if the domain of fff were [[[−−− 101010 ,,, 10]10]10]?
There would be a global maximum at (10, 10
4
e
10
). (And the graph would be restricted to − 10 ≤ x ≤
- Use Newton’s Method with an initial guess of x 0
x
0
x 0
= − 1 to find the next three approxima-
tions to a solution of x
3
x
3
3
- x + 1 = 0. Then test your final approximation to see if it appears
to be close to a root.
Recall that x n+
= x n
f(x n
f
′
(x n
= x n
x
3
n
3 x
2
n
Using a calculator with initial guess x 0
= −1, we get the following.
x 0
x
1
x 2
x 3
Check: (− 0 .68233)
3
- (− 0 .68233) + 1 ≈ 0 .0000526, so our approximation seems to be a good one.
- Circle always, sometimes, or never to make each statement below correct.
(a) If f
′
f(1) = 0
′
f(1) = 0
′
(1) = 0 then fff always /sometimes/never has a critical point at xxx = 1= 1= 1.
A critical point is where f
′
is 0 or undefined.
(b) If f
′
f(2) = 0
′
f(2) = 0
′
(2) = 0 then fff always/ sometimes /never has a local maximum or local minimum
at xxx = 2= 2= 2.
f might instead have a terrace point at x = 2; we need f
′
to change sign at x = 2 in order to
guarantee a local extremum there.
(c) If xxx = 3= 3= 3 is a critical point of fff , then f
′
f(3)
′
f(3)
′
(3) is always/ sometimes /never 0.
It may also be that f
′
(3) is undefined.
(d) If f
′′
f (4) = 0
′′
f(4) = 0
′′
(4) = 0, then fff always/ sometimes /never has an inflection point at xxx = 4= 4= 4.
We need f
′′
to change sign at x = 4 to guarantee an inflection point there.
For example, if f(x) = (x−4)
4
then f
′′
(4) = 0 but f has a local minimum rather than an inflection
point at x = 4.
However, if f(x) = (x − 4)
3
then f
′′
(4) = 0 and f does have an inflection point at x = 4.
Also see what happens at x = 0 in problem 7(d).
(e) If f
f f has a global maximum at x = 5
x = 5 x = 5, then f
′
f
′
f
′
(5) is always/ sometimes /never 0.
f
′
(5) might also be undefined, or x = 5 might be an endpoint of the domain.
(f) If f
′
f
′
f
′
(6) = 0 and f
′′
f
′′
f
′′
(6) = − 2 , then f
f f always /sometimes/never has a local maximum at
x = 6
x = 6 x = 6.
If f is concave down with a horizontal slope at x = 6, then f must have a local maximum there.
(g) If f
′
f
′
f
′
(7) = 0 and f
′′
f
′′
f
′′
(7) = 0, then f
f f always/ sometimes /never has a local extremum at
x = 7
x = 7 x = 7.
This means that the second derivative test is inconclusive, so you need to use a different test.
For example, if f(x) = (x − 7)
4
then f
′
(7) = 0 and f
′′
(7) = 0 and f has a local minimum at
x = 7.
However, if f(x) = (x − 7)
3
then f
′
(7) = 0 and f
′′
(7) = 0 and f has an inflection point but not a
local extremum at x = 7.
- The rate of change of a population PPP (((ttt))) of eels is proportional to the size of the population.
When the population is 40000, it is growing at a rate of 400 eels per year. At time t = 0
t = 0 t = 0,
the population is 10000.
(a) Write a differential equation whose solution is P (t)
P (t) P (t).
Rate of change (P
′
) is (=) proportional to (k) size of population (P ) means P
′
= kP.
What’s the value of k? When P = 40000, we know P
′
= 400. That is, 400 = k · 40000, so k = .01.
Thus, we have P
′
=. 01 P.
(b) Solve your differential equation.
The general solution is P (t) = Ae
. 01 t
What’s the value of A? When t = 0, we know P = 10000. That is, 10000 = Ae
0
= A, so
A = 10000.
Thus, we have P (t) = 10000e
. 01 t
(c) When will the population reach 60000?
60000 = 10000e
. 01 t
6 = e
. 01 t
ln 6 = ln e
. 01 t
Take ln of each side.
ln 6 =. 01 t Recall that ln e
z
= z.
t = 100 ln 6 ≈ 179 .176 years
