Question

derivative
f(x)= 3e^(-7x^2)

Answer 1

chain rule!

Answer 2

use the chain rule to get

f(x) = 3*e^(-7x^2)

f ' (x) = d/dx[ 3*e^(-7x^2) ]

f ' (x) = 3 * d/dx[ e^(-7x^2) ]

f ' (x) = 3e^(-7x^2) * d/dx[ (-7x^2) ]

f ' (x) = 3e^(-7x^2) * (-14x)

f ' (x) = -42x*e^(-7x^2)

Answer 3

to find critical point
f ' (x) = -42x*e^(-7x^2) =0
and it would be zero?

Answer 4

yep because you use the zero product property to get

-42x = 0 or e^(-7x^2) =0

but e^(-7x^2) =0 has no solutions

Answer 5

Great! thanks.
do you know what inflection point ?
I have never heard in the class

Answer 6

it's where the concavity flips

Answer 7

from the same question that you helped me,
would it be zero again?

Answer 8

well you need to find f '' to find the intervals of concavity and the inflection point

Answer 9

oh. I see.

Answer 10

do i use product rule first?

Answer 11

I got
41e^(-7x^2)(14x^2-1)

Answer 12

-1/sqrt(14),1/sqrt(14)

Answer 13

for the points?

Answer 14

I think you meant 42 instead of 41

Answer 15

I got

42*e^(-7x^2)*(14x^2-1)

Answer 16

for f ''