anonymous
  • anonymous
Let f(x) = e^x - e^(3 x). Find all extreme values (if any) of f on the interval 0 <= x <= 1. Determine at which numbers in the interval these values occur. Remember to check for endpoint extrema
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
how do i even start?
anonymous
  • anonymous
after finding the deriv of course
anonymous
  • anonymous
f'(x)= e^x-3e^3x

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More answers

.Sam.
  • .Sam.
I don't think there's any
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.Sam.
  • .Sam.
In interval 0 <= x <= 1
anonymous
  • anonymous
no max values but there are min values
.Sam.
  • .Sam.
You sure there are min values?
.Sam.
  • .Sam.
If you're talking from -1<=x<=0 Then we'll have maximum
anonymous
  • anonymous
im this online math study site and the problem has answers and it has only a min value of 1
anonymous
  • anonymous
max/min values occur at the endpoints on the interval [0, 1]
.Sam.
  • .Sam.
That's weird @satellite73 @Hero @TuringTest @bahrom7893
.Sam.
  • .Sam.
I have to go
bahrom7893
  • bahrom7893
f'(x)=e^x-3e^(3x)
bahrom7893
  • bahrom7893
e^x(1-3e^(2x))=0 1-3e^(2x)=0 e^(2x)=1/3 2x=-ln3 x=-(ln3)/2
anonymous
  • anonymous
your problem asks for extreme values on [0, 1]... according to the graph it is only decreasing on this interval. so the max must occur at x=0 and min at x=1.
bahrom7893
  • bahrom7893
ohh correct dpalnc, i didn't even look at the graph nor read the question properly.
anonymous
  • anonymous
youre right dpalnc
anonymous
  • anonymous
i got that already but what is the crit value for the min ?
anonymous
  • anonymous
if you want the value of the min in [0, 1], it's f(1)...
anonymous
  • anonymous
ok thanks

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