This question is killing me and I need a detailed explanation with work on how to prove this.
Let $x>0$. Prove that if $0a-b$.
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calculus
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0What is the derivative of the left-side with respect to $x$? Is it always positive for positive $x$? – 2017-01-09
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0Please put the problem statement in the question itself, and explain what you have tried so far. – 2017-01-09
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0This is how the question was handed to me. I factored out the a and b respectively but I do not see how that will prove it. – 2017-01-09
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0I found that the derivative is (-abe^-bx) +(abe^-ax) – 2017-01-09
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0So, factoring, we have $ab(e^{-ax}-e^{-bx})$ and we know $ab$ is positive. Is $e^{-ax}-e^{-bx}$ for all positive $x$? If it is positive for all positive $x$, what does that tell us noting that for $x=0$ we have $ae^{-bx}-be^{-ax}=a-b$ – 2017-01-09
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0Which can written $ab(e^{-ax} - e^{-bx})$. What can you say ? – 2017-01-09
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0But wait isn't it -ab((e^-bx) -(e^-ax))? since its there is a -a in the derivative? – 2017-01-09
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0I think I see what you did now JMoravitz you switched the e to proper form thus why you got that sorry it took me a second to realize it. – 2017-01-09
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0And yeah since x>0 which means x is positive then e^{-ax}-e^{-bx} for all x>0. But why equal? – 2017-01-09
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0You guys are awesome thank you so much! :) – 2017-01-09
1 Answers
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$x>0$ and $0
$\implies e^{-ax}-e^{-bx}>0.$
thus the function $f: x\mapsto ae^{-bx}-be^{-ax}$ is strictly increasing at $(0,+\infty)$.
observe that $f(0)=a-b$ and conclude.