(-2(e^-x))sinx=0

Question

Set f''(x)=(-2(e^-x))sinx equal to 0 and solve to find the inflection points of f(x)=(e^-x)sinx

Answer 1

You have an error in your problem statement. For f(x) = e^-xsin(x) the second derivative is -2e^-xcos(x):

Use the product rule: for f(x) = u•v, f '(x) = u'•v + u•v'

So for f(x) = e^-xsin(x),

u = e^-x

u' = -e^-x

v = sin(x)

v' = cos(x)

f '(x) = -e^-xsin(x) + e^-xcos(x) = e^-x(cos(x) - sin(x))

So for f '(x) = e^-x(cos(x) - sin(x)), again using the product rule:

u = e^-x

u' = -e^-x

v = cos(x) - sin(x)

v' = -sin(x) - cos(x)

f ''(x) = -e^-x(cos(x) - sin(x)) + e^-x(-sin(x) - cos(x))

f ''(x) = -e^-xcos(x) + e^-xsin(x) - e^-xsin(x) - e^-xcos(x)

f ''(x) = -2e^-xcos(x)

To find the points of inflection, set f ''(x) equal to zero:

-2e^-xcos(x) = 0 when -2e^-x = 0 or when cos(x) = 0

-2e^-x never equals zero (y = 0 is the horizontal asymptote).

cos(x) = 0 at x = π/2 and every π/2 increment after (and before)