Solve f'(x) = 0: Why is There No Solution?

[nesan]

Homework Statement

http://www.wolframalpha.com/input/?i=((1+/+x)+++2)+/+(3(2x+++lnx))^(2/3)+=+0



Solve the equation f'(x) = 0

f(x) = (ln x + 2x)^(1/3)

The Attempt at a Solution

As you can see, for the function

((1 / x) + 2) / (3(2x + lnx))^(2/3) = 0

I can get a value for x which is - 1 / 2

But my textbook says "no solution"

Why is that?

 

[Mark44]

Homework Statement

http://www.wolframalpha.com/input/?i=((1+/+x)+++2)+/+(3(2x+++lnx))^(2/3)+=+0



Solve the equation f'(x) = 0

f(x) = (ln x + 2x)^(1/3)

The Attempt at a Solution

As you can see, for the function

((1 / x) + 2) / (3(2x + lnx))^(2/3) = 0

I can get a value for x which is - 1 / 2

But my textbook says "no solution"

Why is that?

x must be positive in order for ln(x) to be defined.

 

[nesan]

x must be positive in order for ln(x) to be defined.

Oh wow, I totally missed that. Thank you very much. :)

 

[micromass]

That does not mean that wolfram alpha is wrong however. Wolfram alpha uses a definition of the logarithm which is also defined for negative numbers. This definition uses complex numbers.

However, you likely did not encounter complex numbers and complex logarithms yet, so in your problem x=-1/2 is not a valid solution as the logarithm is not well-defined.

This is a bit the danger of using computer software. Their solutions are not wrong, but they can be different solutions from what you want.