Fundamental Theorem of Calc., Inc./Dec., and concavity

  • Thread starter Qube
  • Start date
  • #1
Qube
Gold Member
468
1

Homework Statement



I am having extreme trouble with the following problems:

http://i.minus.com/iYs6ix6otGtLV.png [Broken]

Homework Equations



For 26:

If the first derivative is positive, then the function is increasing. If the first derivative is negative, then the function is decreasing.

If the second derivative is positive, then the function is concave up. If the second derivative is negative, then the function is concave down.

I tried setting (lnx)/x > 0 but I am having difficulty solving this inequality.

I also tried setting (1-lnx)/x^2 > 0 but I am having difficulty solving this inequality.



For 27:

The constant is a dummy variable, and can be ignored. I used the second fundamental theorem of calculus and substituted in 2x, getting the square root of (4x^2 - 2x). I plugged in 2 and got the square root of 12. The answer key says E. Why am I wrong?
 
Last edited by a moderator:

Answers and Replies

  • #2
21
0
For number 27)

Another way of expressing the second fundamental theorem is:

[itex]\frac{d}{dx}[/itex][itex]\int^{h(x)}_{c}g(t)dt[/itex] = h'(x)*g(h(x))

Just applying the above equation:
h'(x) = 2
g(h(x)) = [itex]\sqrt{(2x)2-(2x)}[/itex] = [itex]\sqrt{4x2 - 2x}[/itex]
g(h(2)) = [itex]\sqrt{12}[/itex]
f'(2) = h'(2)*g(h(2)) = 2[itex]\sqrt{12}[/itex]
 
Last edited:
  • #3
21
0
For number 26)

Firstly, the function is only defined for x > 0.

- In the equation below, since x is always positive and x cannot equal zero, we can multiply through by x:
[itex]\frac{ln(x)}{x}[/itex] > 0
ln(x) > 0
- The natural logarithm is 0 when x = 1 and is larger than 0 (the derivative is positive) when x > 1. Therefore, the function is increasing for x > 1 and decreasing for 0 < x < 1.

- In the equation below, since x2 is always positive and x2 cannot equal zero, we can multiply through by x2:
[itex]\frac{1-ln(x)}{x2}[/itex] > 0
1 - ln(x) > 0
ln(x) < 1
- The natural logarithm is 1 when x = e and is less than 1 (the second derivative is positive) when 0 < x < e. Therefore, the function is concave up for 0 < x < e and concave down for x > e.

If I am unclear in any way, I will be glad to clarify.
 
Last edited:

Related Threads on Fundamental Theorem of Calc., Inc./Dec., and concavity

Replies
15
Views
4K
Replies
2
Views
537
Replies
7
Views
1K
  • Last Post
Replies
1
Views
784
  • Last Post
Replies
2
Views
2K
Replies
7
Views
32K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
978
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
803
Top