Question

f(x) = (ax^2) / (e^ (bx^2)), where a and b are constants.

If f attains an absolute maximum value of 1 at x=2, find the possible values of a and b f(x) = (ax^2) / (e^ (bx^2)), where a and b are constants.

If f attains an absolute maximum value of 1 at x=2, find the possible values of a and b @Mathematics

Answer 1

I found a=(e^(1/2))/2 and b=1/2 but I think I may have done something wrong somewhere

Answer 2

\[ax^{2}/e ^{bx ^{2}}\]

Answer 3

one possible way to do this is to derive f(x) twice wrt x and plug in x = 2


the value of f''(2) will then be negative because the max is the point which is on a concave downward piece of the graph


So f''(2) < 0


You can then use this inequality to solve for either a or b

Answer 4

from what wolfram alpha is spitting out, it looks like it'll be easier to solve for 'a'

Answer 5

I took the ln of the top and bottom of f then took the derivative of f and set it to 0 when x was equal to 2 and then solved for a which I got to be (e^(1/2))/2

Then I plugged that into f when x=2 and f=1

does this sound like it would work to you?

Answer 6

Let's try what you're proposing


when you derive with respect to x you get


\[\Large f'(x) = -2ax e^{-bx^2} (bx^2-1)\]


Plug in x = 2 and f'(x) = 0 and simplify to get


\[\Large 0 = -4ae^{-4b} (4b-1)\]


then solve for a to get a = 0


This sounds valid, but there's a problem: If a = 0, then f(x) = 0 for all x, which would mean that you wouldn't have a max at x = 2 (f(x) = 1 wouldn't even be true as well)


So this means that this method wouldn't work.

Answer 7

that's not the derivative I got I used f=\[\ln(ax)/bx ^{2} \]and got \[(1-2\ln(ax))/bx^3\] for my derivative

Answer 8

I thought it was f(x) = (ax^2) / (e^ (bx^2))


I see how you got the denominator when you took the ln of both numerator and denominator, but I don't see how you got the numerator

Answer 9

so it should be \[\Large f(x) = \frac{\ln(ax^2)}{bx^2}\]

Answer 10

yes that is what I took the derivative of

Answer 11

the derivative of that was (1/x)*(bx^2) - ln(ax)*(2bx) all over (bx^2)^2

Answer 12

then I simplified

Answer 13

hmm the derivatives should come out the same, not sure why they're not though

Answer 14

I think what I did was ok but I'm not completely sure

Answer 15

well I'm using wolfram alpha, so I know it's correct, but your derivative should be equivalent

Answer 16

oh wait, I'm not even thinking clearly here lol


say you have the fraction 1/2, which is nonzero.


Now take the natural log of the numerator and denominator to get


ln(1)/ln(2)

and evaluate


ln(1)/ln(2) = 0/ln(2) = 0


Since 1/2 is not equal to 0, this means that ln(1)/ln(2) is not equivalent to 1/2


This example shows us that even though we applied the natural log to both the numerator and denominator, the number changed completely.

Answer 17

when I put a=(e^(1/2))/2 and b=1/8 into the function and graph it it does have a maximum of 1 at x=2 though

Answer 18

that may be true, but applying the natural log the way you did was invalid (by reasoning shown above)

Answer 19

you can also see the flaw when you compare the graphs of the original f(x) function and the modified function (with the natural logs) and they won't be the same

Answer 20

yeah, so I guess I need to take the derivative of the original function then

Answer 21

you can take the natural log of both sides, but you may have to do implicit differentiation, and then you'll have to convert back anyway

So it's best to tackle it w/ out the use of natural logs

Answer 22

by both sides, I mean the equation: f(x) = (ax^2) / (e^ (bx^2))

Answer 23

yeah that's where I got mixed up

Answer 24

looks like a really annoying derivative to do though, I hope it's solvable when it's done

Answer 25

Thanks for the help looking at this