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STRWABERRY

@STRWABERRY

Beginner

Posted: 15 years ago

#261

ohh thanks please helpp me out iin other q's too =))

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akhl

@akhl

Dazzler

Posted: 15 years ago

#262

Need to think over these as I mostly answer questions in Physics, Chemistry, Mathematics - not in Biology.

I have seen you in yaho answers. :)

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akhl

@akhl

Dazzler

Posted: 15 years ago

#263

Originally posted by: a1989peanut
My question is :

dont forget to show ALL work for full credit: the function is f(x)=x(lnx)^2

What are the first and second derivatives?[/quote]

By product rule of differentiation,

f ' (x) = x d/dx[(lnx)^2] + d/dx (x) * (lnx)^2

Using chain rule of differentiation,

f ' (x) =x * 2 lnx * 1/x + 1 * (lnx)^2

f ' (x= 2 lnx + (lnx)^2

Differentiating again,

f '' (x) =2 * 1/x + 2 lnx * 1/x

f '' (x) = (2/x) (1 + lnx)

Ans: First derivative = 2 lnx + (lnx)^2

Second derivative = (2/x) (1 + lnx)

[quote]a: What is the domain of the f(x)?[/quote]

lnx is valid only if x > 0

Therefore the domain of f(x) is

x > 0

or (0, INFINITY)

[quote]b: where is there a vertical or horizontal asymptote?[/quote]

f ' (x) = 2 lnx + (lnx)^2

There is no finite value of a such that f ' (x) -> infinity when x -> a

Therefore there is no vertical asymptote.

f(x) = x (lnx)^2

When x-> infinity then f(x)-> infinity

Therefore there is no horizontal asymptote.

[quote]c: find all critical numbers[/quote]

For critical numbers

f ' (x) = 0

2 lnx + (lnx)^2 = 0

lnx (2 + lnx) = 0

lnx = 0, -2

x = 0, 1/e^2
x = 0 is not valid as the domain of f(x) is x > 0

Ans: 1/e^2

[quote]d: test whether the function is a maximum or minimum and show your method of choice (first or second derivative test)[/quote]

f ' (x) = 0 at x = 1/e^2

f '' (x) = (2/x) (1 + lnx)

f '' (1/e^2) =2 e^2 [1 + ln(1/e^2)]
= 2 e^2 (1 - 2)

= -2 * e^2 . This is < 0

Therefore, by second derivative test, the function has a maximum at x = 1/e^2

[quote]e: find the point(s) of these local maxima or minima (both x and y coordinates)[/quote]

f (x) = x (lnx)^2

At point of local maximum

x = 1/e^2

Therefore y = f (x)= (1/e^2) [ln(1/e^2)]^2

= 1/e^2 * (-2)^2

= 1/e^2 * 4

= 4/e^2

Ans: Point of local maximum: x = 1/e^2, y = 4/e^2

[quote]f: show the direction of concavity[/quote]

The function has a local maximum. Therefore it is concave down.

Ans: concave down

[quote]g: find the point of inflection (both x and y coordinates)

At point of inflection,

f '' (x) = 0

(2/x) (1 + lnx) = 0

lnx = -1

x = 1/e

f(x) = x (lnx)^2

f(1/e) = (1/e)[ln(1/e)]^2

f(1/e) = (1/e) (-1)^2 = 1/e

Ans: x = 1/e, y = 1/e

Edited by akhl - 15 years ago

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akhl

@akhl

Dazzler

Posted: 15 years ago

#264

It would be better if you post your answer so that I can check. Anyway, here are the answers.

1. find the derivative of the following using on sine and consine functions:

a. cosecant

Let y = csc(x)

y = 1/sin(x)

dy/dx = [-1/sin^2(x)] * cos(x)

dy/dx = (-1/sinx) * (cosx/sinx)

dy/dx = -csc(x) * cot(x)

Answer: - cosecant * cotangent

Can do others like this?

11. Find the specific anti derivatives for the for the velocity and the position:

a. a(t) = -9.8 v(0)=20 s(0) = 100

v(t)= v(0) + integral a(t) dt

= v(0) + integral -9.8 dt

= v(0) - 9.8 t

= 20 - 9.8 t

s(t) = s(0) + integral v(t) dt

= 100 + integral 20 - 9.8 t dt

= 100 + 20 t - 9.8 * 1/2 * t^2

= 100 + 20 t - 4.9 t^2

12. . Find f(x)

a. f'(x) = sin(x) + cos(x)

f(x) = integral sin(x) + cos(x) dx

= cos(x) - sin(x) + C

b. f''(x)=cox(x)

f(x) = integral cos(x) dx

= -sin(x) + C

c. f''(x)= 9.8

f(x) = integral 9.8 dx

= 9.8 x + C

d. f''(x) sinx+cosx, f(0)= 3 f'(0)=4

Check question again. You must have typed something wrong.

_________________.

There are too many. I have answered a few. Later when I find time, I will answer the rest.

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akhl

@akhl

Dazzler

Posted: 15 years ago

#265

10) Find the integral of the given function. Write the integral together with the constant C. Then substitute the x=1 and for F(x), substitute F(1). That will give you value of C

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Heena

@Dream-Land

Navigator

+ 3

Posted: 15 years ago

#266

Hi everyone, Hope you all are doing fine and healthy :-)

I urgently need some help and I really hope someone could be able to help me in here.

I have a question which is related to "POVERTY IN INDIA"

and the question is,

What can be done to fight against poverty in India?

What should the indian priminister do to fight against poverty and to help the poor people in India?

P.S I would be thankful if you can find som sources graph showing figure about the poverty :-)

Hope to get reply from anyone as soon as possible .. Thanks alot :-)

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akhl

@akhl

Dazzler

Posted: 15 years ago

#267

If you search for poverty in India in the following

http://www.ohchr.org

then it may be of help to you

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Heena

@Dream-Land

Navigator

+ 3

Posted: 15 years ago

#268

Originally posted by: akhl
If you search for poverty in India in the following
http://www.ohchr.org

then it may be of help to you

Thanks

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maryam

@aishu_fan

Voyager

Posted: 15 years ago

#269

i have a test in math tomrrow and im having problems with these questions....plz plz help me asap!!! thnk u so much!

write the exponential expression 10x^5/4
same for 9x^9/7

let f(x)= 16+6x and g(x)= 3x-15. find f(g(x)).

for the function f(x)=(2-3x)^2 find f-1. determine whether f-1 is a function.

find the roots of the polynomial equation
6x^4-5x^3-65x^2+85x-21=0 (clueless with this question)

find a quadratic equation with roots -2+4i and -2-4i

find the asymptotes and graph the equatin( u dont have to graph it plz justt ell me how to graph it)

y= -x-5/-4x-4

y= 2x-1/3x+4

points of Discontinuity ?

x+3/ x^2-16x+63

plz thank you!!!!!