Question

Let f(x)=e^x-e^4x
Find all extreme values (if any) of f 0 12 years ago

Answer 1

1st step: Find derivative.

Answer 2

What would that be?

Answer 3

I did that and then set it equal to 0, but couldnt find what x would equal

Answer 4

derivative of e^x is itself.

Answer 5

what did you get after differentiating e^(4x)

Answer 6

i got the whole derivative to be e^x-4e^x and now what do u get when u set that equal to zero?

Answer 7

you are missing something

Answer 8

\[e^x-4e^{4x}=e^x(1-4e^{3x})\]
\[e^x(1-4e^{3x})=0\]
Now set both factors =0

Answer 9

and then what do you get? this is where im stuck

Answer 10

Well is e^x ever 0?

Answer 11

No so thats canceled. Then the other one i got x=-ln(1/4) / 3

Answer 12

but the answer says it should have it in the form of E

Answer 13

(also i don't like to say that cancels; i just there is no solution to that)
\[e^{3x}=\frac{1}{4} => \ln(\frac{1}{4})=3x\]
did i miss something?
how did you get -?

Answer 14

You could do ln(1)-ln(4)=-ln(4)
I can see how you would the negative then
-ln(4)=3x

Answer 15

then divide both sides by 3

x=-3ln(4)
or
x=ln(4^(-3))
or
x=ln(1/64)
or
x=-ln(64)

Answer 16

I see what you did, now im stuck at ln(1/4)=3x

Answer 17

anyways this number doesn't occur on the interval [0,1]

Answer 18

you divide by 3 and then dont you get ln(1/4)/3

Answer 19

or -ln(64)

Answer 20

why

Answer 21

above
i put it step by step using properties of log

Answer 22

ln(a/b)=ln(a)-ln(b)

Answer 23

cln(a)=ln(a^c)

Answer 24

ln(ab)=ln(a)+ln(b)
ln(1)=0

Answer 25

ln(1/4)=ln(1)-ln(4)=0-ln(4)=-ln(4)

1/3*ln(1/4)=1/3* (-ln(4))=-1/3*ln(4)=ln(4^(-1/3))

And I made an error above.

Answer 26

but still by properties of log we can rewrite this

Answer 27

thank you!