Problem of the Week #116 - June 16, 2014

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SUMMARY

The discussion centers on solving the equation $\displaystyle \left(\int f(x)\,dx\right)\left(\int\frac{1}{f(x)}\,dx\right) = -1$. Participants lfdahl and Pranav provided solutions, establishing that if $g(x)=\int f(x)\,dx$, then $g(x)h(x)=-1$ leads to the conclusion that $g(x)=Ae^{\pm x}$, where $A$ is a constant. Consequently, the functions satisfying the original equation are $f(x)=g'(x)=\pm Ae^{\pm x}$.

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Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Find all functions $f$ that satisfy the equation $\displaystyle \left(\int f(x)\,dx\right)\left(\int\frac{1}{f(x)}\,dx\right) = -1$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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This week's problem was correctly answered by lfdahl and Pranav. You can find Pranav's solution below.

[sp]Let $\displaystyle g(x)=\int f(x)\,dx$ and $\displaystyle h(x)=\int \frac{1}{f(x)}\,dx$.

Also, $g'(x)=f(x)$ and $h'(x)=\dfrac{1}{f(x)} \Rightarrow h'(x)=\dfrac{1}{g'(x)}$.

$$g(x)h(x)=-1 \Rightarrow g'(x)h(x)+h'(x)g(x)=0$$
Substitute $h(x)=-\dfrac{1}{g(x)}$ and $h'(x)=\dfrac{1}{g'(x)}$
$$\Rightarrow -\frac{g'(x)}{g(x)}+\frac{g(x)}{g'(x)}=0 \Rightarrow \left(\frac{g'(x)}{g(x)}\right)^2=1 \Rightarrow \frac{g'(x)}{g(x)}=\pm 1 $$
$$\Rightarrow g(x)=Ae^{\pm x}$$
where $A$ is some constant.

$$\boxed{f(x)=g'(x)=\pm Ae^{\pm x}}$$[/sp]
 

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