Mathematical Induction презентация

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:

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

Then Sn is true for all positive integer numbers n.

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,….

We have

Question 6:

Question 5:

e) the open first quadrant

e) I and III

b) a nonconstant function

is positive for all x > 0

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:

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

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.

Hence

We have

that the sequence

In fact,

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

Therefore, there are no points a such that

because at a point like 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

Solution: If f (x1) = x1, then