Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=e^(-x)-e^(-2x) on the interval [0,1]

Question

mathteacher1729 · Answer

Do you know how to use first and second derivatives?

anonymous · Answer

Yes. I know that when the first derivative is equal to 0 that is where you have a critical pt. I just can't figure out how to take the first derivative of the function or find where it's equal to 0

myininaya · Answer

$$y=e^{g(x)} => y'=g'(x)e^{g(x)}$$

anonymous · Answer

WTF??

myininaya · Answer

$$y'=(-x)'e^{-x}-(-2x)'e^{-2x}$$

anonymous · Answer

follow what myininaya said and work out values for x. I obtained 2 solutions.. x=1 and -ln(1/2). One is a minimum the other a maximum. You can use 2nd derivatives to check, or else much simpler put in the values for x and see what y is

anonymous · Answer

*x=-1

anonymous · Answer

ah wait you are only considering the inteval [0,1]

anonymous · Answer

think you lost the 2 on the way, otherwise I agree

myininaya · Answer

yep looks that way lol

anonymous · Answer

thanks all. i see now.

myininaya · Answer

$$y'=(-1)e^{-x}-(-2)e^{-2x}=-e^{-x}+2e^{-2x}=e^{-2x}(-e^{x}+2)=0$$

myininaya · Answer

$$e^{-2x} \neq 0, but -e^{x}+2=0 \text{ has a solution}$$

anonymous · Answer

^^

myininaya · Answer

now you want to find absolute max and min
so just plug in critical numbers on the interval [0,1]
and plug in your endpoints into the original function to see which is the highest and which is the lowest output