Question

Let f and g be differentiable functions with
the following properties:
(i) g(x) > 0 for all x
(ii) f(0) = 2
If h(x) = f(x)g(x) and h′(x) = f(x)g′(x),
then f(x) =

Answer 1

g(x) is always positive, and we know f(x) evaluated when x = 0 is 2. The function, therefore, starts at y-value 2. If we got the derivative of h(x) to be f(x)g'(x) from the product rule, then we must determine there is another term f'(x)g(x), that evaluates to 0. f(x) must be a constant, so f(x) = 2.