Goal: Find the critical point of f\'(x)= 2e^(2x)- e^(-x) Problem: have calculati

Question

Goal: Find the critical point of f'(x)= 2e^(2x)- e^(-x)

Problem: have calculations correct (from board) however DO NOT UNDERSTAND SOME STEPS

- I must not understand my basic algebra/ how to use natural logs of "e"

why does 2e^(2x)=e^(-x) --> 2e^(2x) [e^(2x)] =1

and when do you know to use the natural log? is that just to get rid of the "e"?

LOOKING FOR VERBAL, clear, easy to understand explanation

would love if anyone had a website recommendation, words of wisdom, or a memory trick

thanks so much

Explanation / Answer

I'm trying to read your answer, but when I save it and zoom in enough for it to be legible it gets pretty fuzzy, but here's how I'd explain how to work the problem:

Critical point = inflection point = f'(x)=0

So, take the derivative of e^(2x) + e^(-x) = 0 ==> f'(x) = 2e^(2x) - e^(-x) = 0

This is in a pretty "user-friendly" setup to solve graphically in your calculator, giving a result of x=-0.2310491

Algebriacally (for clarity):

Add e^(-x) to both sides, then take the natural log of both sides and you get:

2x + ln(2) = -x

3x = -ln(2)

x = -ln(2)/3

x = -0.2310491

If you have e^(some power of x) on both sides, taking the natural log of both sides eliminates the "e", and brings the exponents down as your new arguments. The tricky part is remembering your rules of exponents, which you'll get better at with practice. I hope this helps!

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