Solving for f(x)=|log x| in Function Question Homework Statement

[f(x)]

Homework Statement

If f(x)=| log x | , then state whether the following are true or false :
a)y'(1+) =1/x
b)y'(1)=1
c)y'(0)= ∞ (infinity)

Homework Equations

|y| = y if y >=0
|y| = -y if y<0

The Attempt at a Solution

a)The first one is true because if x > 1, then y > 0 and | y | is y. So, y' = 1/x

b)If x=1, y=0 and so |y| = y . So y' =1/x = 1

c)If x=0, y is not defined ?

I am having trouble with b) and c) . I don't understand whether to check by putting values first (which would give all constant, and IMO is wrong) or later as I have done.
The answers are T,F,F

Any help is appreciated

 

[Office_Shredder]

For part b, (I'm assuming y=f), y'(x) exists if y'(x+) = y'(x-) (this is for a particular x). So you need to show the derivative from the left = the derivative from the right (or show it doesn't)

Otherwise, for example, let y=|x|. y(0) = 0, so y'(0) = x' = 1. But this is clearly false.

For part c, I'm tempted to guess they want y'(x) as x->0 from above (otherwise you're absolutely correct). This is actually negative infinity (it cna be seen just by graphing y), so it's false either way.

 

[HallsofIvy]

Homework Statement

If f(x)=| log x | , then state whether the following are true or false :
a)y'(1+) =1/x
b)y'(1)=1
c)y'(0)= ∞ (infinity)

Homework Equations

|y| = y if y >=0
|y| = -y if y<0

The Attempt at a Solution

a)The first one is true because if x > 1, then y > 0 and | y | is y. So, y' = 1/x

b)If x=1, y=0 and so |y| = y . So y' =1/x = 1

c)If x=0, y is not defined ?

I am having trouble with b) and c) . I don't understand whether to check by putting values first (which would give all constant, and IMO is wrong) or later as I have done.
The answers are T,F,F

Any help is appreciated

Office shredder made a good try but what you've written makes very little sense. For one thing you say that f(x)= |log x| but then never mention f again! Is y = f?

"y'(1+)= 1/x" What does "1+" mean? OfficeShreder interpreted it to mean the "right derivative" or "limit from the right" but if that were correct and you are asking for the derivative as you approach x= 1 from the right, then there would be no "x" in the derivative.

If x> 1 then f(x)= log x. What is the derivative of that? What is the limit of the derivative as x-> 1 from above? If x< 1, then f(x)= - log(x). What is the derivative of that? What is the limit of that as x-> -1 from below?

Yes, you are correct that f(0) is not defined and neither is f'(0). I personally don't like saying that something is infinity when it is not defined! However, here, I think you need to look at what happens to |log x| and its derivative for numbers like x= 0.0001, x= 0.00000001, etc.

 

[f(x)]

Thx OfficeShredder and Halls for the help.

Office shredder made a good try but what you've written makes very little sense. For one thing you say that f(x)= |log x| but then never mention f again! Is y = f?

"y'(1+)= 1/x" What does "1+" mean?

sorry about the ambiguity but I can't help, I just copied all that's there in in the text. I rechecked but i see no corrections to make. I s'pose 1+ to mean Numbers greater than one. And about y being f, i think its true else the question doesn't make sense to me.

If x> 1 then f(x)= log x. What is the derivative of that? What is the limit of the derivative as x-> 1 from above? If x< 1, then f(x)= - log(x). What is the derivative of that? What is the limit of that as x-> -1 from below?

LHL is not equal to RHL, so is the function not differentiable at x=1?