Question

challenging ib calc question: http://i58.tinypic.com/296di4z.png

please help! I do not understand the second line "It can be shown that..."

Answer 1

that is what you are being asked to show

Answer 2

@satellite73 but show what? the markscheme just equated the derivative of the first equation to 0, and found the min from there, but i do not understand what the sentence with "it can be shown -equation- where f(n)(x) represents the nth derivative of f(x)"?

Answer 3

Hello, FF:

I'd advise you to start with Part (a). First, let n=1. This results in your needing to find the first derivative of f(x). You could do that either by following the derivative formula given in the problem statement, or by using the product rule in the usual way to find the derivative of \[f(x)=xe ^{2x}. \] Your choice.

In regard to that "it can be shown" business: you are not being asked to do that, but rather to accept the fact that there is a simple formula for the derivative of f(x) that depends upon the value of n. If you WANT to derive that formula, that's another matter, and I'd be glad to guide you through that derivation.

Answer 4

FF: i take it that you are well acquainted with the uses of the first and second derivatives in identifying critical values, possible inflection points, and so on. Those applications are certainly valid in this particular problem that you've posted.

Why not get started? If\[f(x)=xe ^{2x}, \]
please find the first and second derivatives of this function. Either do it in the traditional way (using the product and chain rules), or use the "can be shown" formula in this problem statement.

Answer 5

@mathmale

http://i61.tinypic.com/14mc0mh.jpg

both produced a negative number

Answer 6

I'm going to assume that your two derivatives are correct. Have you taken the first derivative and set it = to 0, to obtain any critical values? If so, what is / what are the critical value (s)?

Answer 7

it is shown in the picture. are the numbers x=-3/2 and -1/2 critical numbers?

Answer 8

By definition, a critical number is an x-value for which the first derivative is zero. Have you checked x=-1/2 in the first derivative to determine whether or not the first deriv. is zero?

same question for x=-3/2.

Answer 9

this is where I'm a little confused; I can't remember if I'm supposed to plug it into the original equation, or the first derivative equation to check

Answer 10

Using the product rule to find the first derivative, I got:\[f'(x)=xe ^{2x}(2)+e ^{2x}(1),\]which can be factored as follows: \[f'(x)=e ^{2x}(1+2x).\]

Answer 11

yes that was what I got

Answer 12

Let's sum up where we are so far. We've both found the first derivative of f(x), right? We got the same result. We set this result, the first derivative, = to 0 and solved for x. I got x=-1/2; you say you got x=-1/2 and x=-3/2.

We both want to know whether x=-1/2 truly is a critical value, and to do that we substitute x=-1/2 into the first derivative (not into f(x) to see whether the first derivative is actually zero at x=-1/2. Right?

Answer 13

ok

Answer 14

So I'm willing to bet my $ that x=-1/2 IS a critical value of f(x).
Now, you claim that x=-3/2 is also a critical value of f(x). To check this, we must substitute x=-3/2 into the first derivative (not into the original function f(x). Would you do that, please? Does the first derivative = 0 at x=-3/2?

Answer 15

I thought we had to substitute x=-3/2 into the second derivative, because I got that answer from setting the 2nd derivative to 0. I checked for the first derivative and it is 0. I also plugged in x=-3/2 into the 2nd derivative and obtained a 0.

Answer 16

\[f'(-3/2)=e ^{-3}(1+2[-3/2])=??\]

Answer 17

If this is zero, x=-3/2 is a critical value; if not zero, x=-3/2 is not a critical value. Your decision?

Answer 18

critical value when i plugged it into 2nd derivative

Answer 19

But, FF, this whole discussion was about the first derivative and any possible critical values.

Answer 20

there is one critical value of f(x), and that c. v. is x=-1/2.

Now (only now) we turn to the second derivative. Why would we do that? Because the 2nd derivative is a powerful tool for determining the direction of concavity of the graph of a function. One such application is to determine whether we have a minimum or a maximum at our critical value, x=-1/2.

Answer 21

To take advantage of this powerful tool, we have to find the 2nd derivative of f(x).
Hav e you done that? If so, what was your result for the 2nd derivative?

Answer 22

2nd derivative of f(x) = e^2x(3+2x)

sorry i do not know how to type the fancy equation stuff

Answer 23

Looks like we'll need to go through finding the 2nd derivative. You've obtained \[f''(x)=(3+2x)e ^{2x}\] and Ihave obtained \[f''(x)=2e ^{2x}(1+1+2x)=4e ^{2x}(1+x)\]

Answer 24

They can't both be right!
How do you want to handle this...coming to an agreement on what the 2nd derivative is in this case?

Answer 25

I have a feeling you are right.. I'm trying to work it out again. see my weakness is the base e. I have been stuck on questions with 'base e' in them..

Answer 26

yep I got the same answer as you now! we can move on to the next step

Answer 27

Now I see where your x=-3/2 came from.
Actually, FF, what is important here is ONLY whether or not the 2nd derivative is positive or negative at the critical value, x=-1/2.

Using your formula for the 2nd derivative, and letting x=-1/2, we get a POSITIVE result. Agree with that? Be sure the math is clear for you.

Using my formula, and letting x=-1/2 again, we get a POSITIVE result again.
So..FF..we don't need to fight over who's correct!

Let's wrap this up:

If we have a critical value and suspect that we have either a min or a max at that c. v., one very fast way of doing that is to substitute that critical value into the SECOND derivative.

Case 1: If the 2nd deriv. is POSITIVE, then the graph is concave upward and we have a minimum at that critical value.

Case 2:

Answer 28

If the 2nd derivative is NEGATIVE, the graph is concave DOWNWARD and we have a MAX at that critical value.

So: our critical value is x=-1/2.

our 2nd derivative is (+) at that x-value.

Therefore, we have a MIN / MAX at x=-1/2. Which one do you think is correct here?

Answer 29

a MIN value? since we plu x=-1/2 into the 2nd derivative, we get a positive number.

Answer 30

Yes! One more time: If substituting a critical value into the SECOND derivative produces a POSITIVE result, we have a MIN at that critical value. If a NEGATIVE result, we have a MAX there. OK?

Answer 31

OK! yay!

Answer 32

Now, FF, f(x) is defined as\[f(x)=xe ^{2x}.\]

Answer 33

ok

Answer 34

We've found that x=-1/2 is a c. v., right? If so, please evaluate f(x) at x=-1/2.

\[f(-1/2) = ?\]

Answer 35

\[f(-1/2)=(-1/2)e ^{2(-1/2)}=?\]

Answer 36

Hint: 2(-1/2)=-1.

Answer 37

-0.1859597206

Answer 38

I just plugged it into my calc?

Answer 39

The reason I'm asking you to do this is that you must give the coordinates of P, the MINIMUM we've just found. The x-coord. is -1/2 and the y-coord about -0.1860. Yes.

So, by finding these coordinates, we've finished (a), right?

Answer 40

yes! oh awkward I thought we moved into the 2nd question

Answer 41

Not yet, but we'll do that right now!

Provided that you want to, of course. :)

Answer 42

yes of course! please proceed!

Answer 43

Inflection point? What's that? Explain it in your own words.

Answer 44

the point where the concavity changes i.e positive concavity changes into neg concavity

Answer 45

or vice versa. Excellent.
And how does one find the x-coord. of an infl. pt.?

Answer 46

take the 2nd derivative and set to 0?

Answer 47

Right, and solve the resulting equation for x. But we're not done yet. We have to determine whether the sign of the 2nd derivative actually changes at that x-value. If it does, then yes, we do have an infl. pt. If it does not change sign, then sorry, we do not have an infl. pt.

Answer 48

ok on it!

Answer 49

You've found what you believe to be the 2nd derivative.
Please take your 2nd derivative and set it = to 0. Then solve the resulting equation for x.

Answer 50

x=-1

Answer 51

So you have completed Part (b) already!

Answer 52

wait that's it? i thought we had to test the point? on a number line?

Answer 53

Actually, you're right and I'm wrong.

Answer 54

Your 2nd deriv. is 0 at x=-1, right?
Please choose a test number from each side of x=-1, e. g., x1=-2 and x2=?

Answer 55

I'm not sure whether to test the number against the original function, 1st derivative, or 2nd derivative. i just tested it on the 2nd derivative for x=-1 and the answer was 0, which I do not think is correct?

Answer 56

Evalute your 2nd deriv at both test numbers. Are the resulting signs the same or diffrerent?

to answer your question, you'd substitute x=-1 into the 2nd derivative to determine whether or not the 2nd deriv. were really zero at x=-1. If yes, then you ahve correctly found the value of x you wanted. Next step is to go thru that test nubmer busienss I've just described.

Answer 57

So, how about evaluating the 2nd deriv. at x=-2 and x=0?
All we care about are the signs...we do not need or want the magnitudes.

Answer 58

sorry disregard my previous statement

Answer 59

Likely something is wrong. We are told to show that there's an inflection point at x=-1, so I'm assuming that x=-1 is correct, and that the 2nd deriv will NOT be zero at values of x other than x=-1.

Answer 60

So:\[f''(-2)= pos ~or~\neg?\]

Answer 61

\[f''(0)=positive~or~negative?\]

Answer 62

sorry please disregard my previous statement. \[concave up = [-1,\infty)\] \[concave down = (-\infty, -1]\]

Answer 63

am i correct?

Answer 64

Cool. So, you've shown that the sign of the 2nd deriv. changes at x=-1, correct? If so, you have then verified that you do indeed have an infl pt. at x=-1. Congrats.

Answer 65

so now it appears that you have done (a), (b) and (c) correctly (although I have not actually checked your results for (c) ).

Answer 66

yay! what about the last 2 steps? im trying to sketch this thing right now, and i have no idea how to induce

Answer 67

You'.ll recall that we found f(-1/2) so that we'd have both coordinates of the minimum, correct?

And that we actually have that point P.
We 'll have the coordinates of point Q, the inflection point, as soon as you've evaluated f(x) at x=-1.

You knw that the curve will be concave up at x=-1/2 and that the direction of concavity changes at x=-1.

Comfortable with this so far?

Answer 68

So it appears that we know a lot about the physical appearance of this function.

Answer 69

yep. i just tried sketching it

Answer 70

One more thing: What does x-intercept mean?

Answer 71

when x=0

Answer 72

Actually, we get an x-intercept when y=0.

Try setting our function f(x) = to 0 and solving for x. Result?

Answer 73

oops I meant that; I got confused -:) when function f(x)=0, result =0

Answer 74

So, in summary:
Infl. pt.: (0, ?)
minimum: (-1/2, -0.1839)
horizontal intercept: (0,0)
You have already identified intervals on which the graph is concave up and down.

Let the fun begin!

Let's graph all this stuff.

\