Review Questions:
1. What is the relationship between the graph of tan(2x) and tan(x)? Write tan(2x) in
terms of sin(x) and cos(x), and calculate limx→π/3 tan(2x).
The graph of tan(2x) is the graph tan(x) compressed horizontally by a factor of 2 (or
2 sin(x) cos(x)
sin(2x)
= cos
stretched by a factor of 1/2). tan(2x) = cos(2x)
2 (x)−sin2 (x) (p.s. there are more than
one ways to write tan(2x) in terms of sin(x) and cos(x)). Because cos function is continuous, 2x is continuous, cos(2x) is continuous, hence limx→π/3 cos(2x) = cos(2π/3) = − 21 6= 0.
√
√
lim
sin(2x)
Similarly, limx→π/3 sin(2x) = 23 , so limx→π/3 tan(2x) = lim x→π/3 cos(2x) = − 3.
x→π/3
2. f (x) = sin x1 is an (a) even function. (b) odd function. (c) increasing function. (d)
decreasing function.
1
= sin(− x1 ) = − sin x1 . It is not even, or increasing, or decreasing.
It is odd because sin −x
3. Show that there is a x such that x2 = ex .
Use Intermediate value theorem. F (t) = t2 − et is continuous, F (−1) = 1 − e−1 > 0,
F (0) = 0 − 1 = −1 < 0, so there must be some x between −1 and 0 such that F (x) = 0
i.e. x2 = ex .
(
b,
x=0
is continuous.
4. Find b such that f (x) =
− 12
e x , x 6= 0
1
1
f is already continuous everywhere other than 0. limx→0 f (x) = limx→0 e− x2 = lim− 1 →−∞ e− x2 =
x2
0. So b = 0 if we want f to be continuous at 0.
Exercises:
1. Write cos(4x) in terms of cos x. How about cos(5x)?
Use cos(2a) = 2 cos2 (a) − 1, we have cos(4x) = 2 cos2 (2x) − 1 = 2(2 cos2 x − 1) − 1 =
8 cos4 x − 8 cos2 x + 1.
Use cos(a + b) = cos a cos b − sin a sin b, we have cos(5x) = cos(4x) cos x − sin(4x) sin x.
Because sin(2a) = 2 sin a cos a, cos(4x) = 8 cos4 x − 8 cos2 x + 1, cos(5x) = cos(4x) cos x −
sin(4x) sin x = (8 cos4 x − 8 cos2 x + 1) cos x − 2 cos(2x) sin(2x) sin x
= 8 cos5 x − 8 cos3 x + cos x − 2(2 cos2 x − 1)(2 cos x sin x) sin x
Because sin2 x = 1−cos2 x, cos(5x) = 8 cos5 x−8 cos3 x+cos x−2(2 cos2 x−1)(2 cos x sin x) sin x =
8 cos5 x − 8 cos3 x + cos x − 4(2 cos2 x − 1) cos x(1 − cos2 x)
1
2
= 16 cos5 x − 20 cos3 x + 5 cos x
2. Find the inverse function of f (x) = ln(x+1)−ln(x−1). What is its domain and range?
x+1
2
The domain of f is (1, ∞). When x is in (1, ∞), f (x) = ln( x+1
x−1 ). Because x−1 = 1+ x−1 is
x+1
+
decreasing on (1, ∞), ln is increasing, f (x) is a decreasing function. As x → 1 , x−1 → ∞,
x+1
x+1
ln x+1
x−1 → ∞, and when x → ∞, x−1 → 1, ln x−1 → 0, so the range of f is (0, ∞). Hence,
the domain of f −1 is (0, ∞) and the range of f −1 is (1, ∞).
y
To calculate f −1 , let y = ln( x+1
x−1 ), then e =
x
f −1 (x) = eex +1
−1 , x > 0.
3. Find the tangent line of y =
√
x+1
x−1 ,
hence ey x − ey = x + 1, x =
ey +1
ey −1 .
So,
x at (1, 1).
The tangent line must pass through (1, 1), so it must be of the form y − 1 = k(x − 1)
where k is the slope.
√
By the definition of the tangent line, k = limx→1
1√
limx→1 √x+
1
=
1
2,
so the tangent line is y = 2x − 1.
√
x− 1
x−1
= limx→1
x−1
√ √
(x−1)( x+ 1)
=