1. Use l’Hˆopital’s Rule to evaluate the limit, if it exists.
(a) (6 points) lim
x→0
tan 3x sin 2x Solution: Since lim
x→0tan 3x = 0, lim
x→0sin 2x = 0, and both tan 3x and sin 2x are differentiable near x= 0,
x→0lim tan 3x sin 2x = lim
x→0
3 sec23x 2 cos 2x = 3
2 by l’Hˆopital’s Rule.
(b) (6 points) lim
x→0 csc x − cot x Solution: Since lim
x→0| csc x| = ∞, lim
x→0| cot x| = ∞, the limit is of the indeterminate form ∞ − ∞.
Hence
x→0lim csc x − cot x = lim
x→0
1 − cos x sin x = lim
x→0
sin x
cos x = 0 by l’Hˆopital’s Rule.
2. (6 points) Use the Squeeze Theorem to prove that
x→0limxcos2 x = 0.
Solution: For any x 6= 0, since
0 ≤
xcos2 x
≤ |x|
and
x→0lim0 = 0 = lim
x→0|x|, we have
x→0lim
xcos2 x
= 0 by the Squeeze Theorem.
Hence
x→0limxcos2 x = 0.
3. (8 points) Let f be the Dirichlet function
f(x) =
(0 if x is rational 1 if x is irrational.
Prove that f is discontinuous at any x ∈ R.
Solution: Take ε = 1 2.
For any x ∈ R = Q ∪ R \ Q and for any δ > 0,
there exist r ∈ Q ∩ (x − δ , x + δ ) and z ∈ R \ Q ∩ (x − δ , x + δ ) such that either
if x ∈ Q =⇒ | f (x) − f (z)| = |0 − 1| = 1 > 1 2= ε
or
if x ∈ R \ Q =⇒ | f (x) − f (r)| = |1 − 0| = 1 > 1 2 = ε.
Either case proves that f is discontinuous at x. Since x is arbitrary, f is discontinuous on R.
4. Suppose that
f(5) = 1, f0(5) = 6, g(5) = −3, and g0(5) = 2.
Find the following values.
(a) (4 points) f g0
(5)
Solution: If both f and g are differentiable at x then, by the Product Rule, f g0
(x) = f0(x) g(x) + f (x) g0(x).
Therefore
f g0
(5) = f0(5) g(5) + f (5) g0(5) = 6 · (−3) + 1 · 2 = −16.
(b) (4 points) f g
0
(5)
Solution: If both f and g are differentiable at x and g(x) 6= 0 then, by the Quotient Rule, f
g
0
(x) = f0(x) g(x) − f (x) g0(x)
g2(x) .
Therefore
f g
0
(5) = f0(5) g(5) − f (5) g0(5)
g2(5) =6 · (−3) − 1 · 2
(−3)2 = −20 9 .
5. (a) (6 points) Find the derivative of y = cos4 sin3x.
Solution: By the Chain Rule, dy
dx = 4 · cos3 sin3x · (−1) · sin sin3x · 3 · sin2x· cos x
= −12 · cos3 sin3x · sin sin3x · sin2x· cos x
(b) (6 points) Find d35
dx35 xsin x.
Solution: For j ≥ 2 and k ≥ 1, since djx
dxj = 0 and d4 jsin x
dx4 j = sin x, and by the binomial formula and the Product Rule, we have
d35
dx35 xsin x = 35dx
dx· d34sin x
dx34 + x ·d35 sin x
dx35 = 35 · d2sin x
dx2 + x ·d3sin x
dx3 = −35 sin x − x cos x.
(c) (6 points) Use implicit differentiation to find an equation of the tangent line to the curve x2/3+ y2/3= 4 at the point (−3√
3, 1).
Solution: By using implicit differentiation, we have 2
3 x−1/3+ y−1/3dy
dx = 0 =⇒dy
dx = − y x
1/3
. Thus, at the point (−3√
3, 1), we have dy
dx = − 1
−3√ 3
1/3
= 1
33/2
1/3
= 1
√3.
Hence the tangent line to the curve x2/3+ y2/3= 4 at the point (−3√
3, 1) has an equation y− 1 = 1
√3(x + 3√ 3).
6. (8 points) Find the absolute maximum and absolute minimum values of f(x) = 3x4− 4x3− 12x2+ 1 on the interval [−2, 3].
Solution: Since
f0(x) = 12x3− 12x2− 24x = 12x (x2− x − 2) = 12x (x + 1) (x − 2), the citical numbers of f in [−2, 3] are x = −1, or 0, or 2. At the critical numbers, we have
f(−1) = 3 + 4 − 12 + 1 = −4, f (0) = 1, f (2) = 48 − 32 − 48 + 1 = −31.
At the endpoints, we have
f(−2) = 48 + 32 − 48 + 1 = 33, f (3) = 243 − 108 − 108 + 1 = 28.
Hence
x∈[−2,3]max f(x) = f (−2) = 33 and min
x∈[−2,3]f(x) = f (2) = −31.
7. (a) (6 points) Let f be continuous at c and assume that f (c) > 0. Prove that there exists δ > 0 such that f (x) > 0 for all x ∈ (c − δ , c + δ ).
Solution: Since f (c) > 0 =⇒ f(c)2 > 0.
By taking ε = f(c)2 , the ε − δ definition of the continuity of f at c implies that there exists a δ > 0, such that
if x ∈ (c − δ , c + δ ) then | f (x) − f (c)| < f(c)2
=⇒ −f(c)2 < f (x) − f (c)
=⇒ f (x) > f(c)2 > 0.
(b) (6 points) Let f be differentiable at c. Prove that f is continuous at c.
Solution: Since f be differentiable at c,
x→clim f0(c) = f(x) − f (c)
x− c exists, and
x→clim| f (x) − f (c)| = lim
x→c
f(x) − f (c)
x− c · (x − c)
=
limx→c
f(x) − f (c) x− c · lim
x→c(x − c)
= f0(c) · 0 = 0.
Hence
limx→c f(x) = f (c) and f is continuous at c.
8. Let f be differentiable on (a, b).
(a) (6 points) Suppose that f0(x) 6= 0 for all x ∈ (a, b). Prove that f is 1 − 1 on (a, b).
Solution: For any x16= x2 ∈ (a, b), there exits c lying between x1 and x2, by the Mean Value Theorem, such that
f(x1) − f (x2) = f0(c)(x1− x2) 6= 0 =⇒ f (x1) 6= f (x2) =⇒ f is 1 − 1 on (a, b).
(b) (6 points) Suppose that f0(x) > 0 for all x ∈ (a, b). Prove that f is increasing on (a, b).
Solution: For any x1< x2∈ (a, b), there exits c ∈ (x1, x2), by the Mean Value Theorem, such that
f(x1) − f (x2) = f0(c)(x1− x2) < 0 =⇒ f (x1) < f (x2) =⇒ f is increasing on (a, b).
9. (16 points) For x 6= ±1, let f (x) = 2x2 x2− 1.
(a) Find the intervals on which the graph of y = f is increasing or decreasing, critical numbers and local extreme values, if exist.
Solution: Since
f0(x) = (x2− 1) (4x) − 2x2· 2x
(x2− 1)2 = −4x (x2− 1)2
(> 0 if x ∈ (−∞, −1) ∪ (−1, 0),
< 0 if x ∈ (0, 1) ∪ (1, ∞),
f is increasing on (−∞, −1) ∪ (−1, 0), f is decreasing on (0, 1) ∪ (1, ∞), x= 0 is a critical number of f ,
f(0) = 0 is a local maximum value of f .
(b) Find the intervals on which the graph of y = f is concave upward or downward.
Solution: Since
f00(x) = (x2− 1)2(−4) + 4x · 2(x2− 1)2x
(x2− 1)4 = 12x2+ 4 (x2− 1)3
(> 0 if x ∈ (−∞, −1) ∪ (1, ∞),
< 0 if x ∈ (−1, 1),
the graph of f is concave upward on (−∞, −1) ∪ (1, ∞), the graph of f is concave downward on (−1, 1).
(c) Find the asymptotes, if it exists, of the graph y = f . Solution: Since
x→±∞lim 2x2
x2− 1 = 2 =⇒ y = 2 is the horizontal asymptoe, and since
2x2
x2− 1 >2x2
x2 = 2 for |x| > 1 =⇒ the graph of f lies above y = 2 for |x| > 1 Since
x→−1lim± 2x2
x2− 1 = ∓∞ and lim
x→1±
2x2
x2− 1= ±∞ =⇒ x = ±1 are vertical asymptotes of the graph of f .
(d) Sketch the graph of f . Solution:
(−∞, −1) (−1, 0) (0, 1) (1, ∞)
f0 +ve +ve −ve −ve
f00 +ve −ve −ve +ve
f % concave upward % concave downward & concave downward & concave upward
