To prove the continuity of the function, can we determine that the function is continuous in the interval as long as we prove the continuity at two endpoints (in the open and closed interval)?

To prove the continuity of the function, can we determine that the function is continuous in the interval as long as we prove the continuity at two endpoints (in the open and closed interval)?

When limf (x) = f (x0) x - > x0, f is said to be continuous at x0
After introducing the concept of increment, the definition of continuity is equivalent to Lim △ y = 0 △ X - > 0 (that is, when the change of X is very small, the change of Y is 0)
Or use ε-δ Narrative method: if any ε> 0, exists δ> 0, properly | x-x0|

If the function is second-order differentiable in an interval, can it be proved that the function is continuous? Why?

Yes, because the premise of derivability (except generalized derivability) under [general definition] is continuity. You can write it in definition and know that derivability must be continuous

It is proved that if f (x) is differentiable on interval I and the derivative function is bounded on I, then f (x) is uniformly continuous on I

Let | f '(x) | ≤ M
Then, for any x, y ∈ I, according to the Lagrange mean value theorem, there is | f (y) – f (x) |≤ m | Y-X |
So, Ren Gei ε> 0, take δ=ε/ M. Then when Y-X < δ When there is | f (y) – f (x) |≤ m | Y-X | < M( ε/ M)= ε
When the proposition is proved, it is proved

It is proved that the function f (x) = 1-1 / X is an increasing function on (negative infinity, 0)

Let X10, the original function is an increasing function from negative infinity to 0

It is proved that the function f (x) = (2-x) / (x + 2) is a subtractive function on (negative 2, positive infinity)

Take a and B at (- 2, + ∞) and set A0
So f (a) - f (b) > 0
So f (a) > F (b)
So f (x) is a subtractive function on (- 2, + ∞)

If f (x) is a bounded real function, f (x + 13 / 42) + F (x) = f (x + 1 / 6) + F (x + 1 / 7) find the minimum positive period of F (x)

It is well proved that 1 / 7 is the period of the function
There is a symmetry in the condition
Similarly, it can be concluded that 1 / 6 is the period of the function
Then the reciprocal of their least common multiple is also the period of the function
That is, the minimum period that can be determined is 1 / 42
Another condition is needed for a smaller cycle
I think so