Question

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

Answer 1

how do i even start?

Answer 2

after finding the deriv of course

Answer 3

f'(x)= e^x-3e^3x

Answer 4

I don't think there's any

Answer 5

In interval 0 <= x <= 1

Answer 6

no max values but there are min values

Answer 7

You sure there are min values?

Answer 8

If you're talking from -1<=x<=0
Then we'll have maximum

Answer 9

im this online math study site and the problem has answers and it has only a min value of 1

Answer 10

max/min values occur at the endpoints on the interval [0, 1]

Answer 11

That's weird
@satellite73 @Hero @TuringTest @bahrom7893

Answer 12

I have to go

Answer 13

f'(x)=e^x-3e^(3x)

Answer 14

e^x(1-3e^(2x))=0
1-3e^(2x)=0
e^(2x)=1/3
2x=-ln3
x=-(ln3)/2

Answer 15

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.

Answer 16

ohh correct dpalnc, i didn't even look at the graph nor read the question properly.

Answer 17

youre right dpalnc

Answer 18

i got that already but what is the crit value for the min ?

Answer 19

if you want the value of the min in [0, 1], it's f(1)...

Answer 20

ok thanks