Question

Critical points of \(f(x)=e^{-x} sinx\) on \([0, 2 \pi]\)

Answer 1

I got \[\large f'(x)=e^{-x}(cosx-sinx)\]
\[\large 0=e^{-x}(cosx-sinx)\]?

Answer 2

how did you differentiate e^-x * sin x?

Answer 3

\[\LARGE f'(x)=e^{-x} cosx-sinx e^{-x}\]
\[\LARGE =e^{-x}(cosx-sinx)\]

Answer 4

\[\large f'(x)=(e^{-x})(sinx)'-(sinx)(e^{-x})'\]

Answer 5

cool
then obtain the points between the interval given at which the value or tangent is zero

Answer 6

Well I know you can't do \(\large e^{-x}=0\) but \(cosx-sinx=0\)?

Answer 7

e^-x by itself, no solution exist. meaning that it will never be zero

Answer 8

\[\large cosx=sinx\]?

Answer 9

yes?

Answer 10

OH, now I see what you mean.. took me awhile. Thanks nin!

Answer 11

trig can be confusing and isn't very intuitive to find critical points
you did the right thing by factoring, because that is when you can easily determine zeros