3 Calc Questions

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Guest

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1. A function is defined by f(x) = xe^(-2x) with domain (0,10)
a) find all x values where it is increasing and all values where it is decreasing ---> derivative is e^(-2x)-2xe^(2x)
b) give absolut min and max

2. Let f be the function defined by f(x) = x^3+ax²+bx+c and having the following properties
(i) graph has a point of inflection at (0,2)
(ii) mean value of f(x) on [0,2] is -3

a) determine a, b and c
b) determine the value of x that satisfies the conclusion of the mean value therorem for f on [0,3]

3. position func x(t)=sin(pi*t²)
a) velocity ---> cos(pi*t²)(2pi*t)
b) Acceleration ---> 2pi[cos(pi*t²)-tsin(pi*t²)]
c) Wheat values of t does the particle change direction ---> sign chart, decreasing [-1,0) increasing (0,1] so t=0 is where is changes direction
d) find what values for t for which it is moving to the left
 

soroban

Elite Member
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Jan 28, 2005
Messages
5,588
Hello, calcuman!

Here's #1 . . .

1. A function is defined by f(x) = xe<sup>-2x</sup> with domain (0,10)
a) Find all x values where it is increasing and all values where it is decreasing
b) Give absolute min and max
You found the derivative: . f '(x) = e<sup>-2x</sup>(1 - 2x)

(a) f(x) is increasing when f '(x) is positive.
. . . When is: .e<sup>-2x</sup>(1 - 2x) > 0 ?

Since e<sup>-2x</sup> is always positive, we can divide by it
. . . and we have: . 1 - 2x > 0

Solve for x: . x < 1/2

Hence, f(x) is increasing on the interval [0, 1/2)
. . . and it is decreasing on the interval (1/2, 10]


(b) x = 1/2 is a critical value.
. . . To the left, function is increasing: /
. . . At x = 1/2, the tangent is horizontal: ---
. . . To the right, it is decreasing: \

The curve looks like this: . /<sup>----</sup>\ . . . Hence, it is a maximum point.
Since all the other points are "lower", it is an <u>absolute</u> maximim: (1/2, 1/[2e]).


For the absolute minimum, check out the endpoints of the interval.
. . . At x = 0: . f(0) = 0
. . . At x = 10: . f(10) = 10·e<sup>-20</sup> = 0.00000021
Therefore, the absolute minimum is at (0,0).
 

ChaoticLlama

Junior Member
Joined
Dec 11, 2004
Messages
199
First you made one typo: e^(-2x)-2xe^(-2x)

1. a)
0 = e^(-2x) - 2xe^(-2x)
e^(-2x) = 2xe^(-2x)

this equation is true when x = 1/2

b) Because the function is increasing to the left of x = 1/2 and decreasing to the right of it, then absolute max is:
y(1/2) = 1/(2e)

You must check the endpoints, so look at the domain [0,10]
x = 0
y(0) = 0

x = 10
y(10) = 2.06115 * 10^(-8)

Therefore (0,0) is absolute min
 
G

Guest

Guest
thanks, after i posted i worked out that problem, and i got the answers, but they didnt make sense - i see that missed negative messed me up - any help on the 2 or 3?
 

ChaoticLlama

Junior Member
Joined
Dec 11, 2004
Messages
199
3. x(t) = sin(pi*t²)

a) v(t) = cos(pi*t²)(2pi*t)

b) a(t) = 2pi[cos(pi*t²) - 2*pi*t²*sin(pi*t²)]

c) The particle changes direction when v(t) changes from positive to negative. Therefore, one must find the zeros.

v(t) = cos(pi*t²)(2pi*t)

This function equals zero when either cos(pi*t²) = 0 or when (2pi*t) = 0

cos(pi*t²) = 0
pi*t² = pi/2
t² = 1/2
t = ±√1/2

(2pi*t) = 0
t = 0

Therefore the particle changes direction 3 times at t = ±√1/2 and t = 0

d) the particle is moving left when v(t) > 0

Do a sign chart around the 3 zeros to find that the particle is moving left when t < -√1/2 and [0,√1/2]
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, calcuman!

2. Let f be the function defined by f(x) = x<sup>3</sup> + ax<sup>2</sup> + bx + c and having the following properties

(i) graph has a point of inflection at (0,2)
(ii) mean value of f(x) on [0,2] is -3

a) determine a, b and c
b) determine the value of x that satisfies the conclusion of the mean value therorem for f on [0,3]
"Inflection point at (0,2)" tells us two things: . f ''(0) = 0 . and . f(0) = 2

We have: . f '(x) = 3x2 + 2ax + b
. . . Then: . f ''(x) = 6x + 2a

Since f ''(0) = 0, we have: . 6(0) + 2a = 0 . ---> . a = 0

. . . The function (so far) is: . f(x) .= .x<sup>3</sup> + bx + c

Since f(0) =2, we have: . 0<sup>3</sup> + b(0) + c = 2 . ---> . c = 2

. . . The function is: . f(x) .= .x<sup>3</sup> + bx + 2


"The mean value of f(x) on [0,2] is -3" means:
. . . If we evaluate the intergral of f(x) from 0 to 2
. . . and divide by the width of the interval: 2 - 0 = 2,
. . . we will get -3 (the average value of the function).
. . . . . . . . . .<sub>2</sub>
Integrate: . ∫ (x<sup>3</sup> + bx + 2)dx . = . x<sup>4</sup>/4 + bx<sup>2</sup>/2 + 2x
. . . . . . . . . .<sup>0</sup>
. . . = . [16/4 + 4b/2 + 4] - [0 + 0 + 0] . = . 2b + 8

. . . Divide by 2: . b + 4

. . . And this equals -3: . b + 4 = -3 . ---> . b = -7

Therefore: . f(x) .= .x<sup>3</sup> - 7x + 2

I'll let <u>you</u> do part (b).
 
G

Guest

Guest
for that last post, we didnt do integration yet, so idk how to do it that way - how do you do it without integration?
 
G

Guest

Guest
i dont really get what moving left means, can you explain please

and do you mean it is moving left when x(t) >0, because for c i did a sign chart for v(t), and where the sign chart shows positive, that is where x(t) is increasing. but are you saying to do a double prime chart, to show where v is increasing - but wouldnt that be concavity.
on my sign chart, i have x= 0 rel min and x = +- 1/rad2 as rel max. inc over the (-1,-1/rad2)U(0,1/rad2) and dec (-1/rad2,0)U(1/rad2,1). but that would be where x(t) >0 not v(t)
confused...
 
G

Guest

Guest
bump
 
G

Guest

Guest
bump, i need help with #2
 
E

elchichita

Guest
#2:

f(x) = x^3+ax²+bx+c which means that

f'(x) = 3x^2+2ax+b and f''(x) = 6x+2a...we want an inflection point at x=0, so f''(0)=2a=0, which implies a=0. So we come back to the def of f to get

f(x) = x^3+bx+c......But we want (0,2) an inflection point, then f(0)=c=2.
Therefore

f(x) = x^3+bx+2. We'll use the last piece of information to get b.

Mean Value =1/2 int_0^2 (x^3+bx+2) dx=1/2 [x^4/4+bx^2/2+2x]_0^2=
1/2[4+2b+2]=3+b=-3, so b=-6

Summary

f(x)=x^3-6x+2
 
G

Guest

Guest
again, how do you do that last part without an integral ? i dont know how that works yet. and what about part b

b) determine the value of x that satisfies the conclusion of the mean value therorem for f on [0,3]
 

pka

Elite Member
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Jan 29, 2005
Messages
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Use the MEAN VALUE theorem: [f(2)-f(0)]/[2-0] = -3. From this we get 4+b=-3 or b=-7.

On [0,3], [f(3)-f(0)]/[3-0]=2, so for what α is it true that f′(α)=2? That is: you solve, 3(α)<SUP>2</SUP>-7=2.
 

tangents

New member
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May 11, 2005
Messages
9
part b of that question is basically asking the area under the curve? isnt there a way to do itusing limits prior to integrals? Im not sure exactly how its done but i had seen a similar quesiton worded like that
 

pka

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Jan 29, 2005
Messages
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Tangents, elchichita, and all,
This question has nothing whatsoever to do with integrals or area!
The mean is not the average value in this case.
Often the word ‘mean’ is sometimes average but not for functions on closed inteverals.
 

tangents

New member
Joined
May 11, 2005
Messages
9
alright i got that you actually use the mean value theorem and that b=-7
however for part b, using the mean vlaue theorm from [0,3] you get 2 which you showed. but now why did you make the derivative equal to 2? doesnt it just say whats the value of x when you use the mean value theorem from [0,3]?
 
G

Guest

Guest
going back to 3d for a sec guys,

this is what someone posted
d) the particle is moving left when v(t) > 0

Do a sign chart around the 3 zeros to find that the particle is moving left when t < -√1/2 and [0,√1/2]

i am confused here, because the sign chart taht i had done for the equation for v(t), did show those things - but wouldnt that mean that as the points when the slope is increasing, which are correct as he stated them, that x(t) would be increasing. so therefore, should i be doing a sign chart for x double prime, to find where v(t) is equal to zero, or should it just be moving left where x(t) > 0 - and also a clarifcation - what is moving left?
 
G

Guest

Guest
and also, so would the answer to 2b be x = +- rad3
 
G

Guest

Guest
bump
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,321

What happens when the velocity is zero? Has the particle changed directions?
Look at the graph. The red trace is the position, the blue is the velocity.
The velocity is zero for t=√[(2k-1)/2], k is positive integer.
(The value of t≥0 because this a position problem.)

Now from the graph, our particle begins at 0 ‘moves’ to right 1
then starts left back to –1: then goes to 1 again.
Therefore, can you see that it moves left when v(t) is negative?
Between t=√[(1)/2] and t=√[(3)/2] and then moves right.
This pattern is repeated.
 
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