Question

I have the derivative e^-x^2*(6-12x^2)
how would I set it equal to zero and get the x value

Answer 1

forget the \(e^{-x^2}\) part, that is never zero

Answer 2

set
\[6-12x^2=0\] solve it three easy steps

Answer 3

so i carry the 6 over to make it -6 and then what?

Answer 4

add \(12x^2\) divide by \(12\) take the square root, don't forget the \(\pm\)

Answer 5

oh so the sqrt of 6/12?

Answer 6

aka \[\pm\frac{1}{\sqrt{2}}\]

Answer 7

ohh oki thank you so much

Answer 8

but what happens to the e^-x^2

Answer 9

you just want this to be zero right?

Answer 10

\(e^x\) is never zero, so ignore that factor

Answer 11

i want to get the x value

Answer 12

so i am able to find the value that will maximize an area of a rectangle

Answer 13

you just look at \[\pm\frac{ 1 }{ 2 }\] . these are your critical numbers so test on a number line to see which one is a max and which one is a min. since you are maximizing, you need the max which I found to be positive 1/2

Answer 14

oh thaaank you i see