Find the absolute max and min as well as all the values of x where they occur using the given function and domain.

X=X+e^-5X

Question

Find the absolute max and min as well as all the values of x where they occur using the given function and domain.

X=X+e^-5X

anonymous · Answer

i assume the left hand side is $f(x)$

anonymous · Answer

The derivative I got from this is (1+e^-5x) (-5)

anonymous · Answer

as in $$f(x)=x+e^{-5x}$$

anonymous · Answer

Yes, sorry.

anonymous · Answer

your derivative is wrong i think

anonymous · Answer

should just be 
$$f'(x)=1-5e^{-5x}$$

anonymous · Answer

How do I solve that to find the critical points? How do I make the e go away?

anonymous · Answer

lol

anonymous · Answer

set 
$$1-5e^{-5x}=0$$

anonymous · Answer

then go to 
$$5e^{-5x}=1$$ divide by $5$ and get 
$$e^{-5x}=\frac{1}{5}$$

anonymous · Answer

now get rid of the $e$ by rewriting in logarithmic form as 
$$-5x=\ln(\frac{1}{5})$$

anonymous · Answer

or if you prefer 
$$-5x=\ln(.2)$$
then divide and get 
$$x=-\frac{\ln(.2)}{5}$$

anonymous · Answer

that is how you make the $e$ go away, by taking the log

anonymous · Answer

Thank you! You've been such a big help. I always thought though that on equations that had e in it, like the one we just did, that you still took the derivative of the x. Lol.

anonymous · Answer

you do if you need to 
you have to use the chain rule, so for example if 
$$f(x)=e^{x^2}$$ then 
$$f'(x)=2xe^{x^2}$$

anonymous · Answer

yw

anonymous · Answer

But not when it's connected to something like 1+e to the whatever?

anonymous · Answer

i am not exactly sure what you mean...

anonymous · Answer

I figured it out. Lol