Louis Reasoned and his friend, Noah Wei, are having a conversation. Louis: Clear

Question

Louis Reasoned and his friend, Noah Wei, are having a conversation. Louis: Clearly the product of 2 strictly increasing functions is strictly increasing, so ... Noah: Wait! Is that true? Why is that true? Louis: If 2 things are getting bigger, then their product is getting bigger. Noah: Really? Is that always, always true? Louis: Yes. In fact, I have a proof, provided we assume we are talking about "nice" functions. Let's assume both f and g are differentiable functions on some internal I, and both f > 0 and g' > 0 on I. Then, ... uh oh! What proof strategy did Louis have in mind to show fg is strictly increasing? At what point does the proof strategy fail? That failed step gives a clue that you can use to construct a simple counterexample So, give an interval I and 2 differentiable functions f, g: I rightarrow R, with f' > 0 and g' > 0 on I, but for which fg is strictly decreasing on I.

Explanation / Answer

Let f and g are strictly increasing, and f'>0 and g'>0

Suppose h = f*g

so h' = f'g*g'f

As f and g are strictly increasing and f' and g'>0, we can easily conclude that h'>0 and h is strictly increasing by definition of strictly increasing function.

Suppose one of the function is strictly increasing and other one is strictly decreasing.

Let f is strictly increasing and g is strictly decreasing then also product h = f*g is strictly increasing.

Because, h' = f'g+g'f

but f is strictly increasing.

so addition must become positive.

(b) If f'>0 then only we can say f is strictly increasing. otherwise strategy fail.

(C) If f = (1/x) and g(1/(1+x)), then product f*g becomes strictly decreasing function on the interval [0, infinity).

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