Asked by Anonymous
let f and g be differentiable functinos witht the following properties:
g(x)>o for all x
f(0)=1
if h(x)=f(x)g(x) and h'(x)=f(x)g'(x), then f(x)
multiple choice:
a) f'(x) b)g(x) c) e^x d)0 e)1
it can't be a or b. it also can't be c right? because if f(x) is e^x, then h'(x) would be equal to f(x)g'(x) + g(x)f'(x) but it's not.
so is it either zero or one
h(x) = f(x)g(x) --->
h'(x) = f'(x)g(x) + f(x)g'(x)
if h'(x)=f(x)g'(x), then that means that:
f'(x)g(x) = 0
Since g(x) > 0 for all x, that means that:
f'(x) = 0 for all x.
this means that f(x) is constant. Because f(0) = 1, this implies that
f(x) = 1.
g(x)>o for all x
f(0)=1
if h(x)=f(x)g(x) and h'(x)=f(x)g'(x), then f(x)
multiple choice:
a) f'(x) b)g(x) c) e^x d)0 e)1
it can't be a or b. it also can't be c right? because if f(x) is e^x, then h'(x) would be equal to f(x)g'(x) + g(x)f'(x) but it's not.
so is it either zero or one
h(x) = f(x)g(x) --->
h'(x) = f'(x)g(x) + f(x)g'(x)
if h'(x)=f(x)g'(x), then that means that:
f'(x)g(x) = 0
Since g(x) > 0 for all x, that means that:
f'(x) = 0 for all x.
this means that f(x) is constant. Because f(0) = 1, this implies that
f(x) = 1.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.