Question #36e9f

1 Answer
Sep 17, 2017

We are given;

f(x)<=0
g(x) > 0
h(x) = f(x)/g(x)

The first states that f(x) is negative or zero, the second that g(x) is strictly positive, and the third establishes a new function, h(x) which is a ratio of the first two. We are also told that both are differentiable and increasing.

If a function is increasing, then we can infer that its derivative function must be positive. This can be written as;

f'(x) > 0
g'(x) > 0

In order to show that h(x) is increasing, we must show that h'(x) is also positive. We can find h'(x) using the quotient rule.

h'(x)= (f'(x)g(x) - f(x)g'(x))/(g^2(x))

With the information we have, only one term is potentially negative, f(x), and f(x) appears only once, in the subtracted term in the numerator. When subtracting a negative, the sign becomes positive, so the numerator is therefore positive. The denominator must also be positive, as g(x) is strictly positive, so h'(x) is positive. Notice that f(x) can also be 0, in which case, h'(x) is still positive. h(x) is therefore increasing.

Of course, we already made the assumption that h(x) is differentiable. Returning to h'(x), the denominator term is strictly positive, so we don't have to worry about any 0's there. Furthermore, we are told that f(x) and g(x) are differentiable, so they must also be continuous, and f'(x) and g'(x) exist. Therefore, h'(x) exists, so h(x) is differentiable.

Lastly, h(x) is indeed negative for all points where it is not 0.
The numerator f(x) is negative or 0, and the denominator, g(x), is strictly positive, and a ratio of a positive and a negative is always negative.