Mathematical Induction

[−1,1], and f 2(x) = x 2 for each x from the interval [−1,1].

How many functions like that exist?

Solution. The function f (x) = 0 only if x = 0.
At all other points either f (x) = x or f (x) = – x.
Since the function f is continuous, it obtains values of the same sign on each of the intervals [−1,0) and (0,1].
Therefore, there are exactly four possible cases:

and f 2(x) = x 2 for each x from the interval [−1,1].

How many functions like that exist if it is known that x = ½ is the only point where f is not continuous?

numbers n.
Suppose that
1. S1 is true.
2. If Sk is true, then Sk +1 is also true.

Then Sn is true for all positive integer numbers n.

1. The formula is correct in the

Step 2. Let as assume that the formula is

Our aim is to show, that in this case the formula is also correct for n = k + 1.

case n = 1, because

correct for n = k:

for any n = 1,2,3,….

is correct for any n.

We have

3:

Question 6:

Question 5:

e) the open first quadrant

e) I and III

b) a nonconstant function

is positive for all x > 0

1. The formula is correct in the

case n = 1, because

for any n = 1,2,3,….

And hence

Step 2. Let as assume that the formula is

correct for n = k:

is correct for any n.

Our aim is to show, if our formula is correct for n = k, then it is also correct for n = k + 1.

derive explicit formulae for the derivatives of arcsin x, arccos x, arctan x, and arccot x.

Solution. Using the formula

we obtain the following formula for the derivative of arcsin x:

The range of arcsin x is the interval

Therefore cos(arcsin x) is always non-negative.

x.

Hence

to show that the sequence xn is fundamental:

We have

that the sequence

to some limit L.

In fact,

the derivative is a continuous function and f (f (x)) = x for any x. Furthermore, let f (0) = 1, and f (1) = 0.
a) Is it possible that there exists a number a such that

Solution: Differentiate the equation f (f (x)) = x to obtain

Therefore, there are no points a such that

because at a point like that

that

Solution: The Mean Value Theorem tells us that

Therefore, there are no points a such that

for some point c: 0 < c < 1.

otherwise there would be a point b somewhere between c and a such that

= x. Find

Solution: If f (x1) = x1, then