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Comments
f(x) and g(x) are increasing; therefore, f′(x)>0 and g′(x)>0 for all x∈I. The slope of f(x)g(x), denoted (f(x)g(x))′, is equal to f′(x)g(x)+g′(x)f(x). A "necessary" condition, as @Calvin Lin says, would be that at least one of f(x) and g(x) be positive for all x∈I. A "sufficient" condition to say that f(x)g(x) is always increasing would be to say that f(x)>0 and g(x)>0 for all x∈I.
The statement is false. A counterexample is when f(x)=x and g(x)=x+1. Then f(x)g(x)=x2+x and (f(x)g(x))′=2x+1. Over the range (∞,−21), both f(x) and g(x) are increasing, but f(x)g(x) is decreasing.
if f and g both are greater than 0 then its true .. for the interval I .. else it needs to be checked .. rightly pointed out by aditi .. where y=x <0 for x<0 and e^x is always positive .. !! xe^x then have a critical point atx=-1 which is its minima .. !!
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
f(x) and g(x) are increasing; therefore, f′(x)>0 and g′(x)>0 for all x∈I. The slope of f(x)g(x), denoted (f(x)g(x))′, is equal to f′(x)g(x)+g′(x)f(x). A "necessary" condition, as @Calvin Lin says, would be that at least one of f(x) and g(x) be positive for all x∈I. A "sufficient" condition to say that f(x)g(x) is always increasing would be to say that f(x)>0 and g(x)>0 for all x∈I.
The statement is false. A counterexample is when f(x)=x and g(x)=x+1. Then f(x)g(x)=x2+x and (f(x)g(x))′=2x+1. Over the range (∞,−21), both f(x) and g(x) are increasing, but f(x)g(x) is decreasing.
if f and g both are greater than 0 then its true .. for the interval I .. else it needs to be checked .. rightly pointed out by aditi .. where y=x <0 for x<0 and e^x is always positive .. !! xe^x then have a critical point atx=-1 which is its minima .. !!
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Can you state a sufficient condition for fg to be increasing?
Can you state a necessary condition for fg to be increasing?
That's false. An example: f(x)=x and g(x)=e^x Both are increasing but there product is a non-monotonic function.