Suppose f(x)=x^3+5. find the graph of f(1/3x)

1. If f(2) = 5, f(3) = 2, f(4) = 5, g(2) = 6, g(3) = 2 and g(4) = 0,

find (f · g)(2) + f

g(3)

.

Solution. (f · g)(2) + f

g(3)

= f(2) · g(2) + f(2) = 5 · 6 + 5 = 35.

2. Evaluate the following limit

limx→0

2 −

√

4 − x

2

x

2

.

Solution.

limx→0

2 −

√

4 − x

2

x

2

= limx→0

2 −

√

4 − x

2

x

2

·

2 + √

4 − x

2

2 + √

4 − x

2

= limx→0

4 − (4 − x

2

)

x

2

(2 + √

4 − x

2

)

= limx→0

1

2 + √

4 − x

2

=

1

4

.

3. For which value of the constant c is the function f(x) continuous

on (−∞,∞)?

f(x) = (

c

2x − c x ≤ 1

cx − x x > 1.

Solution. The partial functions of f(x) are continuous for x < 1

and x > 1 because they are polynomials. To get f(x) continuous on

(−∞, ∞) we need

lim

x→1−

f(x) = lim

x→1+

f(x) = f(1).

This happens when c

2 − c = c − 1. Rearranging gives 0 = c

2 − 2c + 1 =

(c − 1)2

, and thus c = 1.

4. Compute

lim

x→π/2+

tan x.

Solution. From the graph of y = tan x, the limit is −∞. Or, since

tan x =

sin x

cos x

and sin(π/2) = 1 and cos(π/2) = 0, tan x has a vertical

asymptote at x = π/2. Thus the limit is either ∞ or −∞. For π/2 <

x < π, we have sin x > 0 and cos x < 0. Thus for x near π/2 but

greater than π/2, tan x < 0. Therefore the answer must be