lurflurf said:Dick's approach is popular, but I never liked it. I do not think this problem merits the use of the derivative. I like to show inductively that
|f(x)-f(y)|<=(x-y)^2 implies
|f(x)-f(y)|<=(x-y)^2/2^n for any natural number n from which the result is obvious.
hint
|f(x)-f(y)|=|[f(x)-f((x+y)/2]+[f((x+y)/2)-f(y)]|<=|f(x)-f((x+y)/2|+|f((x+y)/2)-f(y)|
A constant function is a mathematical function where the output value is the same for every input value. In other words, the function does not change or vary with different input values.
A constant function is different from other types of functions because it always produces the same output value regardless of the input value. Other types of functions may have different output values for different input values.
To prove that a function is constant, you must show that the output value is the same for every input value. This can be done by evaluating the function for different input values and showing that the output remains constant.
Proving that a function is constant can be useful in many mathematical applications. It can help simplify equations and make it easier to solve problems involving the function. It can also provide insights into the behavior of the function and its relationship to other functions.
No, a function cannot be both constant and non-constant. A function is either constant, meaning it produces the same output for every input, or it is non-constant, meaning it produces different outputs for different inputs. It cannot be both at the same time.