Math 113 Hw 5 Worksheet With Answers printable pdf download

Math 113 Hw 5 Worksheet With Answers

Math 113 HW #5 Solutions

1. Exercise 2.5.46. Suppose f is continuous on [1, 5] and the only solutions of the equation

f (x) = 6 are x = 1 and x = 4. If f (2) = 8, explain why f (3) > 6.

Answer: Suppose we had that f (3)

6. Then either f (3) = 6 or f (3) < 6. The ﬁrst is

clearly impossible, because we know f (x) = 6 only when x = 1, 4.

On the other hand, if f (3) < 6, then we have that f is continuous on [2, 3] and f (2) = 8 >

6 > f (3). Therefore, by the Intermediate Value Theorem, there exists a number c between 2

and 3 such that f (c) = 6. However, this is again impossible because c is between 2 and 3 and

so cannot be equal to 1 or 4.

Hence, since f (3)

6 is impossible, we have to conclude that f (3) > 6, as desired.

2. Exercise 2.6.6. Sketch the graph of an example of a function f that satisﬁes all of the

conditions

lim

f (x) =

lim

f (x) =

lim f (x) = 1,

lim f (x) = 1.

Answer:

-5

-4

-3

-2

-1

-2

-3

3. Exercise 2.6.18. Find the limit

lim

+ 4y

Answer: Dividing both numerator and denominator by y

, we get

lim

= lim

(5y

+ 4y)

5 +

Math 113 Hw 5 Worksheet With Answers