Question

Let f(x) = xe^−x2/2 .
i. Find all local maxima and minima of f(x).
ii. Determine the intervals on which f(x) is decreasing.
iii. Determine the intervals on which f(x) is increasing.
Any assistance is appreciated,
Kind regards

Answer 1

Well, can you find the derivative first?

Answer 2

1 moment

Answer 3

the function is a little ambiguous... is it \(\large f(x)=xe^{-\frac{x^2}{2}} \) ????

Answer 4

Agreed.

Answer 5

yep...thats it

Answer 6

where is the ambiguity?

Answer 7

You didn't use any parentheses or notation. The function could have gone a number of ways.

Answer 8

sorry...the one that dpa posted is correct...i'll be more careful next time

Answer 9

so I factor out the x

Answer 10

sorry...I should use the product rule

Answer 11

Anyway to cut a long story short

Answer 12

the derivative is y=(1-x^2).e((-x^2)/2)

Answer 13

Do you know how to find the minima and maxima using the derivative?

Answer 14

Could you advise me please . Do I need a 2nd derivative as well?

Answer 15

You don't have to, no. But you know that there is a minima or maxima where the derivative is 0. Here, there are only two.

Answer 16

\[y''=x(x^2-3)e-^{x ^{2}/2}\]

Answer 17

okay..how do you tell that there are only two ?

Answer 18

okay I plugged the function into my calulator and I get two 0s for the function.
One where x is -1, the other where x is 1. So these two points are the maximum and minimium points...Is that correct....

Answer 19

Sorry, yes that is correct. You'll see that the x^2 - 1 term factors into (x-1)(x+1).
Here the exponential term has no zero, it will never cross the x-axis.

Answer 20

That is, the term with 'e', in it.

Answer 21

Thank you for that tip....!
How do i determine where the intervals of f(x) is increasing or decreasing?

Answer 22

You can set up some simple test values. Here, you can check the -1. Try plugging -2 and 0 into the original (non-differentiated) equation. For the 1, try 0 (which you already have), and 2.

Answer 23

okay