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.
