Please help
How many functions satisfy
f(1)=10 and |f(x)-f(y)| = |x-y|
for all x,y€R?
Answers
Answer:
The first condition merely states that the graph passes through or includes the point (1, 10), so we can ignore that until the end of our analysis. Note that every function f will satisfy the second condition whenever x = y, so we may assume x and y are distinct (thus allowing division by x - y). Now, dividing both sides in the second condition by |x-y| yields a result equivalent to saying that the absolute value of every possible difference quotient between two points [with abscissa values x and y] always equals 1. The difference quotient (or DQ for short) here is (f(y)-f(x))/(y-x). So, the absolute value of the slope of the secant line joining any two points (x, f(x)) and (y, f(y)) is 1. If we temporarily ignore absolute value, this would lead to a line of slope 1, as all possible secants would have slope
Step-by-step explanation:
second stated condition of equality, as f(x) - f(y) = (x + b) - (y + b) = x - y, from which taking absolute values yields the required result. From that, f(1) = 1 + b = 10, so b = 9, resulting in f(x) = x + 9. This is one solution.
However, if the DQ always is -1, then that would also satisfy the second condition, owing to the absolute values. In that case, we infer f(x) = -x + b, and plugging in the point (1, 10) gives f(1) = -1 + b = 10, so b = 11, resulting in f(x) = -x + 11. This is a second solution.