f(x) = e^6x + e^−x Find increasing and decreasing intervals. I understand how to do these questions, but I get really confused when they do it with the e
ok how u find increasing interval??
I need to find the derivative and then do a number line
u know the derivative of e^ something??
So far I got up to =6e^6x-e^-e
I dont know what to do after that
what u do to find increasing or decreasing interval in general ??
after finding derivative
I found the critical numbers and then made a number line and saw where it was increasing and decreasing
I set the first derivative = 0
\[6e^{6x}-e^{-x}=0\]
The function is increasing when its first derivative is ---- The function is decreasing when its first derivative is ---- know what comes in those blanks ?
add e^-x on both side
would I then get e^-x (6e^5x-1)=0 ?
hartnn said right first we need to assure it is greater than 0 or less than 0 after that we will find the interval by puttin it equal to zero
\[6e^{6x}-e^{-x}+e^{-x}=e^{-x}\]
what u will get ??
6e^6x=e^-x
yeah ...u know?? x^(-a)=1/x^a it can also written like this also
@sheetalvee
oh okay, I didnt know that
okay leave ...it ..take natural log on both side
okay. so 6x log 6e = -x log 6 ?
\[\log_e (6*e^6x)=\log_e e^{-x}\]
log 6+ 6x*log e= -x log e
when number and base are equal the its value =1 like here base= e and number =e
log6+6x*1=-x*1 tell me what u got
log6=-7x ?
log_e 6=1.79
1.79=-7x
divide both side by 7
i prefer to take inequalities from beginning only increasing when \(6 e^{6x} > e^{-x} \) decreasing when \(6 e^{6x} < e^{-x} \)
yes thatz is the best way..i carried bcoz she said she put =0
wait how did you get 1.79
u got the crictical point now
we took natural log so natural log of 6 = 1.79
also by now you should be able to solve it like this : |dw:1414954148408:dw| without using logs if you can't, then better learn it now :)
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