Question

At what point does the graph f(x) = x^2e^-x^2 have a horizontal tangent line ?????? Have no idea :(

Answer 1

What defines a horizontal tangent? The derivative (or slope) is 0, i.e. horizontal.

Answer 2

Can you calculate the derivative of the function?

Answer 3

LOL I can try just one moment

Answer 4

\[f(x)=x^2e^{-x^2}\]
Have I interpreted this correctly?

Answer 5

yes that is right :)

Answer 6

Ok. Any progress on the derivative?

Answer 7

\[2x \times e ^{-x^2}+x^2 \times e ^{-x^2} \times -2x\]

Answer 8

Yep and now to solve f'(x) = 0. Start with factorising.

Answer 9

I have never factorised anything like this :(

Answer 10

Start off by cleaning up.
\[2xe^{-x^2}-2x^3e^{-x^2}\]
That should help.

Answer 11

Would I take out the common factor of \[2xe ^{-x^2}\]

Answer 12

Yep that would work.

Answer 13

\[2xe ^{-x^2} (1-x^2)\]

Answer 14

Correct!

Answer 15

So now, f'(x)=0.

Answer 16

0,1 ?

Answer 17

Clearly the exponential function never equals 0, so that leaves
x = 0 and 1 - x^2 = 0

Answer 18

0, 1 and you missed an answer

Answer 19

pardon?

Answer 20

1 - x^2 = 0 has two solutions for x

Answer 21

+ or - 1

Answer 22

That's is. And those three solutions combined gives your answer.

Answer 23

Excellent thank you so very much :)

Answer 24

Your welcome!

Answer 25

so do I say that the function will be horizontal when x = 0, 1 or -1?

Answer 26

You do indeed.

Answer 27

Thank you again :)