VCE Methods Question Thread!

[j.wang]

what does strictly decreasing have to do with the inequality sign 

[Alicia23]

Hey guys I'm having trouble with these questions:
1. Given that f(1)=2 and f'(x)=4, find the derivative of f(x)ln(x) when x=1
2. Find f'(x): f(x)= ln(x) / x+1 , x>0
3. Find f'(pi): f(x)= 3x^2+1/ cos x

[zhen]

what does strictly decreasing have to do with the inequality sign 

Here's a picture I found on google showing what increasing and strictly increasing means, which might help your understanding of strictly decreasing, since it's just the opposite.

[j.wang]

Here's a picture I found on google showing what increasing and strictly increasing means, which might help your understanding of strictly decreasing, since it's just the opposite.

thanks zhen I think i understand the image, but I'm not sure how the textbook is using strictly decreasing to justify switching the inequality sign in the attached pic?

[zhen]

thanks zhen I think i understand the image, but I'm not sure how the textbook is using strictly decreasing to justify switching the inequality sign in the attached pic?

I'm not sure what you're asking. Could you attach a picture of the question or the part of the textbook you don't understand.

[j.wang]

Oh my gosh silly me... forgot to attach the pic

can you tell that vce is doing my head in 

not sure about the bottom few lines where the textbook explains the alternative method to solve the problem

[zhen]

Oh my gosh silly me... forgot to attach the pic

can you tell that vce is doing my head in 

not sure about the bottom few lines where the textbook explains the alternative method to solve the problem

Basically since the number is between 0 and 1, it's strictly decreasing, as if you square a positive number less than 1 it becomes smaller. Also, since the smaller the number the larger the value (from the fact that it's strictly decreasing). Then x is less than or equal to the value you get.

[MattBro]

Hey,
Could someone please help me differentiate this: cos^3(2x−π/4)

Thanks in advance

[Sine]

Hey,
Could someone please help me differentiate this: cos^3(2x−π/4)

Thanks in advance

Chain rule first let
y =  cos^3(2x−π/4)
u = cos(2x - π/4)   y = u^3
du/dx = -2sin(2x -  π/4)
dy/du = 3u^2
dy/dx = dy/du * du/dx[/tex]
dy/dx = 3u^2 *-2sin(2x -  π/4) [/tex]
dy/dx = 3cos(2x - π/4)^2 *-2sin(2x -  π/4) [/tex]
dy/dx = -6cos(2x - π/4)^2 * sin(2x -  π/4)

I would be hesitant on giving this as my final answer since I introduced the y myself so write out .
d/dx(cos^3(2x−π/4)) = -6cos(2x - π/4)^2 * sin(2x -  π/4)

[ringring]

Hey! Can anyone pls help me with the question in the attached section

Thanks

[zhen]

Hey! Can anyone pls help me with the question in the attached section

Thanks

The coordinates are (2,8) and (2+h,(2+h)³)
So using the gradient formula of a line, the gradient would be

So I got D as the answer, but I may be wrong, so check with someone else.

[LPadlan]

2loge(x+5)=6 solve for x

[Syndicate]

2loge(x+5)=6 solve for x

2loge(x+5)=6
loge(x+5) = 3
x+5 = e^3
x=e^3-5

First post!

http://imgur.com/a/ATUFC
For Q4a, is my answer okay? The answer in the book has exclusive brackets for all of it but you should be able to include the non-infinite end points right?

Tyy

No, you can only include the points that are defined, not the points which aren't (cause it is being approached, not touched).

[LPadlan]

8e^-x-e^x=2 solve for x

[ringring]

How do I solve the attached question?