Natural Logs+Calc related question

[JenniferBlanco]

Hi,

Can someone please help me out and guide me?

Homework Statement

http://img167.imageshack.us/img167/3892/dscn8321rn1.jpg

Homework Equations

Shown above

The Attempt at a Solution

a) f(x) = -2 +ln(x)^2
f'(x)= 2/x
Ans: All real numbers except for 0

b) -2 +ln (x)^2 =0
ln(x)^2= 2
e^2=x^2
Ans: x = +-e

c) I have no idea how to do this. Do I have to differentiate?-Jen

 

[Mathdope]

Hi,
c) I have no idea how to do this. Do I have to differentiate?

How would you find the "slope of a line that is tangent to the graph"? Once you have the slope, what point would lie on the tangent line if it is tangent to the graph? Once you have the slope, and a point, how do you get the equation of the line?

 

[JenniferBlanco]

How would you find the "slope of a line that is tangent to the graph"? Once you have the slope, what point would lie on the tangent line if it is tangent to the graph? Once you have the slope, and a point, how do you get the equation of the line?

OK, so I differentiated the equation and got 2/x
Then I plugged 1 into x, so our slope is 2 while the y value is --> -2 + 0 = -2

and the equation i got is --> y+2=2(x-1)

is that right? Also, are my first two parts right?

-Jen

 

[dynamicsolo]

Looks like you're all set on this problem!

 

[JenniferBlanco]

Thanks Mathdope and dynamicsolo!

 

[HallsofIvy]

b) -2 +ln (x)^2 =0
ln(x)^2= 2
e^2=x^2
Ans: x = +-e

This part is wrong. If the problem were ln(x²)= 2 then taking the exponential of both sides would give you x²= e². However, the problem is [ln(x)]²= 2. You need to take the square root of both sides first, then the exponential: ln(x)= $\pm\sqrt{2}$ so x= $e^\sqrt{2}$ or x= $e^{-\sqrt{2}}$.

 

[Mathdope]

Halls, the OP seemed to be the problem you mention but the original image that she uploaded was the other one, so I think it's ok. The poster just needs a bit more care in where parentheses go.

 

[dynamicsolo]

Halls, the OP seemed to be the problem you mention but the original image that she uploaded was the other one, so I think it's ok. The poster just needs a bit more care in where parentheses go.

Jeez, I didn't even look that closely at what was typed. I was looking at the attachment, so I based my assessment of OP's answers on that version of the problem. (I should be watching what students are typing, but since their answers were correct for the correct statement of the equation, I thought it reasonable that they knew what they were doing...)