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Math 105: Review for Final Exam, Part I - SOLUTIONS
- Consider the function f(x) =
5 − 2 x
f(x) =
5 − 2 x
f(x) =
5 − 2 x
(a) Is this function continuous on the interval (−∞, ∞)
(−∞, ∞)? Explain.
No. The function is discontinuous at x = 2.5, where f is undefined (and has a vertical asymptote).
(b) Compute the average rate of change of f
f f on [2, 2 .01]
[2, 2 .01]
[2, 2 .01].
f(2.01) − f(2)
[
]
(c) Using the limit definition of the derivative, compute f
′
(x)
f
′
(x) f
′
(x).
f
′
(x) = lim
h→ 0
f(x + h) − f(x)
h
provided this limit exists
= lim
h→ 0
3
5 −2(x+h)
3
5 − 2 x
h
= lim
h→ 0
3(5− 2 x)
[5−2(x+h)](5− 2 x)
3[5−2(x+h)]
[5−2(x+h)](5− 2 x)
h
common denominator
= lim
h→ 0
15 − 6 x − (15 − 6 x − 6 h)
[5 − 2(x + h)](5 − 2 x)h
= lim
h→ 0
6 h
[5 − 2(x + h)](5 − 2 x)h
= lim
h→ 0
[5 − 2(x + h)](5 − 2 x)
(5 − 2 x)
2
(d) Find the equation of the tangent line to f
f f at x = 2
x = 2 x = 2.
We want y = mx + b. m = f
′
2
= 6, so y = 6x + b.
[Note that this slope agrees well with our answer from (b) above.]
When x = 2, y = f(2) =
Thus, 3 = 6 · 2 + b, so b = −9 and we have y = 6x − 9.
- Given that f(0) = 2, g(0) = 3, f
′
(0) = 5, g
′
f(0) = 2, g(0) = 3, f
′
(0) = 5, g
′
f(0) = 2, g(0) = 3, f
′
(0) = 5, g
′
(0) = 7, and f
′
(3) = π
f
′
(3) = π f
′
(3) = π compute the following.
(a) h
′
h(0)
′
h(0)
′
(0) if hhh(((zzz) =) =) = fff(((zzz)))ggg(((zzz)))
h
′
(0) = f
′
(0)g(0) + f(0)g
′
(b) j
′
j(0)
′
j(0)
′
(0) if j(z) =
f(z)
g(z)
j(z) =
f(z)
g(z)
j(z) =
f(z)
g(z)
j
′
f
′
(0)g(0) − f(0)g
′
[g(0)
2
]
2
(c) k
′
k
′
k
′
(0) if k(z) = f(g(z))
k(z) = f(g(z)) k(z) = f(g(z))
k
′
(0) = f
′
(g(0)) · g
′
(0) = f
′
(3) · (7) = (π)(7) = 7π
- (a) Find
dy
dt
dy
dt
dy
dt
if y = t
5
t
5
t
t
5
t
- ln (5t) + arctan (5t) + ln(5) + sin 5
y = t
5
t
5
t
t
5
t
- ln (5t) + arctan (5t) + ln(5) + sin 5 y = t
5
t
5
t
t
5
t
- ln (5t) + arctan (5t) + ln(5) + sin 5.
dy
dt
= 5t
4
t
− 5 t
− 2
t
− 6 / 5
5 t
1 + (5t)
2
= 5t
4
t
t
2
t
6 / 5
t
1 + 25t
2
(b) Find
dy
dx
dy
dx
dy
dx
if y =
3
x cos(7x
3
y =
3
x cos(7x
3
y =
3
x cos(7x
3
dy
dx
x
− 2 / 3
cos(7x
3
3
x(− sin(7x
3
)(21x
2
cos(7x
3
3 x
2 / 3
− 21 x
7 / 3
sin(7x
3
(c) Find
dy
dz
dy
dz
dy
dz
if y =
e
z
π
tan 4 − 7 z
y =
e
z
π
tan 4 − 7 z
y =
e
z
π
tan 4 − 7 z
dy
dx
e
z
(tan 4 − 7 z) − (−7)(e
z
π
(tan 4 − 7 z)
2
(d) Find
dy
dr
dy
dr
dy
dr
if y = tan (e
r
2
arcsin(5r)
y = tan (e
r
2
arcsin(5r)
y = tan (e
r
2
arcsin(5r)
dy
dr
= sec
2
(e
r
2
arcsin(5r)
) · e
r
2
arcsin(5r)
[
r
2
1 − 25 r
2
· 5 + 2r arcsin(5r)
]
(e) Find
dy
dx
dy
dx
dy
dx
if y
3
2
2
= 3y
2
y
3
2
2
= 3y
2
y
3
2
2
= 3y
2
. Here we use implicit differentiation.
3 y
2
dy
dx
dy
dx
x
2
dy
dx
3 y
2
dy
dx
dy
dx
x
2
− 6 y
dy
dx
= − 2 xy − 2 x
dy
dx
(3y
2
2
− 6 y) = − 2 xy − 2 x
dy
dx
− 2 xy − 2 x
3 y
2
2
− 6 y
(f) Find
dy
dx
dy
dx
dy
dx
if y = (1 + x
6
8 x
y = (1 + x
6
8 x
y = (1 + x
6
8 x
. Since we have x in the base and the exponent, we need logarithmic
differentiation.
ln y = 8x ln(1 + x
6
y
dy
dx
= 8 · ln(1 + x
6
) + 8x ·
1 + x
6
· 6 x
5
dy
dx
[
8 · ln(1 + x
6
48 x
6
1 + x
6
]
· y
dy
dx
[
8 · ln(1 + x
6
48 x
6
1 + x
6
]
· (1 + x
6
8 x
- Is y = 7e
3 t
y = 7e
3 t
y = 7e
3 t
a solution to the differential equation y
′′
′
− 15 y = 0
y
′′
′
− 15 y = 0 y
′′
′
− 15 y = 0? Explain.
A given function y will be a solution to the differential equation if, when we substitute in y
′′
, y
′
, and
y, the equation is satisfied (that is, both sides of it are equal).
Since y = 7e
3 t
, we know that y
′
= 21e
3 t
and y
′′
= 63e
3 t
from the Chain Rule.
Now we check to see whether our y satisfies the differential equation.
y
′′
′
− 15 y
?
63 e
3 t
3 t
− 15 · 7 e
3 t
?
63 e
3 t
3 t
− 105 e
3 t
?
So, we see that y = 7e
3 x
is in fact a solution to this differential equation.
- Rewrite sin(arctan(5sin(arctan(5sin(arctan(5xxx)))))) as an algebraic expression. [Students in the 8:00, 9:30, and 1:
sections may omit this problem.]
Let θ = arctan(5x). That is, θ is the angle whose tangent is 5x.
We draw a triangle for which
opposite
adjacent
5 x
= 5x.
5 x
z
θ
2
2
= z
2
⇒ z =
1 + 25x
2
sin(arctan(5x)) = sin θ =
opposite
hypotenuse
5 x
1 + 25x
2
- 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 form ∞/∞.
(a) lim
x→∞
x
2
ln x
lim
x→∞
x
2
ln x
lim
x→∞
x
2
ln x
♥ lim
x→∞
2 x
1 /x
= lim
x→∞
2 x
2
(b) lim
z→ 0
sin (12z) − 12 z
z
3
lim
z→ 0
sin (12z) − 12 z
z
3
lim
z→ 0
sin (12z) − 12 z
z
3
F lim
z→ 0
12 cos(12z) − 12
3 z
2
F lim
z→ 0
−144 sin(12z)
6 z
F lim
z→ 0
−1728 cos(12z)
(c) lim
x→ 0
e
x
cos x
lim
x→ 0
e
x
cos x
lim
x→ 0
e
x
cos x
(d) lim
r→ 2
r
3
r − 2
lim
r→ 2
r
3
r − 2
lim
r→ 2
r
3
r − 2
F lim
r→ 2
3 r
2
(e) lim
x→ 0
x
3
lim ln x [Students in the 8:00 and 9:30 sections may omit this problem.]
x→ 0
x
3
lim ln x [Students in the 8:00 and 9:30 sections may omit this problem.]
x→ 0
x
3
ln x [Students in the 8:00 and 9:30 sections may omit this problem.]
This is of the form 0 · (−∞), so we rewrite it as a fraction to turn it into a L’Hopital’s Rule
problem.
lim
x→ 0
x
3
ln x = lim
x→ 0
ln x
1
x
3
♥ lim
x→ 0
1
x
− 3
x
4
= lim
x→ 0
x
x
4
= lim
x→ 0
x
3
