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

[Qube]

Homework Statement

I am having extreme trouble with the following problems:

http://i.minus.com/iYs6ix6otGtLV.png

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?

 

[AliAhmed]

For number 27)

Another way of expressing the second fundamental theorem is:

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

Just applying the above equation:
h'(x) = 2
g(h(x)) = \sqrt{(2x)²-(2x)} = \sqrt{4x² - 2x}
g(h(2)) = \sqrt{12}
f'(2) = h'(2)*g(h(2)) = 2\sqrt{12}

 

[AliAhmed]

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:
\frac{ln(x)}{x} > 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:
\frac{1-ln(x)}{x²} > 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.