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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 domain (−∞, ∞)
(−∞, ∞)? Explain.
No. f 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.
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
′
h
′
(0) if h(x) = f(x)g(x)
h(x) = f(x)g(x) h(x) = f(x)g(x)
h
′
(0) = f
′
(0)g(0) + f(0)g
′
(b) j
′
j
′
j
′
(0) if j(x) =
f(x)
g(x)
j(x) =
f(x)
g(x)
j(x) =
f(x)
g(x)
j
′
f
′
(0)g(0) − f(0)g
′
[g(0)
2
]
2
(c) k
′
k
′
k
′
(0) if k(x) = f(g(x))
k(x) = f(g(x)) k(x) = f(g(x))
k
′
(0) = f
′
(g(0)) · g
′
(0) = f
′
(3) · (7) = (π)(7) = 7π
- Compute dy/dx
dy/dx dy/dx for each of the following.
(a) y = x
5
x
5
x
x
- ln (5x) + arctan (5x) + ln(5) + sin 5
y = x
5
x
5
x
x
- ln (5x) + arctan (5x) + ln(5) + sin 5 y = x
5
x
5
x
x
- ln (5x) + arctan (5x) + ln(5) + sin 5
dy
dx
= 5x
4
+(ln 5)
x
− 5 x
− 2
5 x
1 + (5x)
2
·5+0+0 = 5x
4
+(ln 5)
x
x
2
x
1 + 25x
2
(b) y =
3
x cos(7x
3
y =
3
x cos(7x
3
y =
3
x cos(7x
3
y =
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) y =
e
x
π
tan 4 − 7 x
y =
e
x
π
tan 4 − 7 x
y =
e
x
π
tan 4 − 7 x
dy
dx
e
x
(tan 4 − 7 x) − (−7)(e
x
π
(tan 4 − 7 x)
2
(d) y = tan (e
x
2
arcsin(5x)
y = tan (e
x
2
arcsin(5x)
y = tan (e
x
2
arcsin(5x)
dy
dx
= sec
2
(e
x
2
arcsin(5x)
) · e
x
2
arcsin(5x)
[
x
2
1 − 25 x
2
· 5 + 2x arcsin(5x)
]
(e) y
3
2
2
= 3y
2
y
3
2
2
= 3y
2
y
3
2
2
= 3y
2
[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
- Consider the differential equation y
′
y = − 3 y
′
y= − 3 y
′
= − 3 y.
(a) For what value(s) of CCC and kkk is y = Ce
−kx
y = Ce
−kx
y = Ce
−kx
a solution to this differential equation?
If y = Ce
−kx
, then y
′
= −kCe
−kx
. Therefore, to satisfy the differential equation given, the
following must be true.
y
′
= − 3 y
−kCe
−kx
= − 3 Ce
−kx
This is true when k = 3 and C is any real number.
(b) Find the solution that passes through (1(1(1,,, 5)5)5).
We plug x = 1 and y = 5 into y = Ce
− 3 x
5 = Ce
− 3
⇒ C = 5e
3
, so the solution is y = 5e
3
e
− 3 x
or y = 5e
3 − 3 x
- Find all possible antiderivatives of the following.
(a) g
′
(t) = e
5
5
5 t
g
′
(t) = e
5
5
5 t
g
′
(t) = e
5
5
5 t
Any antiderivative will have the form e
5
t +
t
6
e
5 t
+ C.
(b) h
′
(r) = 3 sin(2r) +
3
r
h
′
(r) = 3 sin(2r) +
3
r h
′
(r) = 3 sin(2r) +
3
r
Any antiderivative will have the form −
3 cos(2r)
r
4 / 3
+ C.
- Evaluate the following limits.
Note that throughout this solution, the symbol ♥ will stand for the phrase “has the form ‘
’ and so,
by L’Hopital’s Rule, is equal to” while the symbol F will stand for the phrase “has the form ‘
’ and
so, by L’Hopital’s Rule, is equal to”.
(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
x→ 0
sin (12x) − 12 x
x
3
lim
x→ 0
sin (12x) − 12 x
x
3
lim
x→ 0
sin (12x) − 12 x
x
3
F lim
x→ 0
12 cos(12x) − 12
3 x
2
F lim
x→ 0
−144 sin(12x)
6 x
F lim
x→ 0
−1728 cos(12x)
(c) lim
x→ 0
e
x
cos x
lim
x→ 0
e
x
cos x
lim
x→ 0
e
x
cos x
(d) lim
x→ 5
35 − 7 x
2 x − 10
lim
x→ 5
35 − 7 x
2 x − 10
lim
x→ 5
35 − 7 x
2 x − 10
F lim
x→ 5
(e) lim
x→ 0
−
x
lim
x→ 0
−
x
lim
x→ 0
−
x
tiny, tiny negative
= large negative
(f) lim
x→ 0
x
lim
x→ 0
x
lim
x→ 0
x
does not exist lim
x→ 0
−
x
= −∞ but lim
x→ 0
x
