Locate any relative extrema and inflection points.
y = x - lnx

Question

Locate any relative extrema and inflection points. 
y = x - lnx

anonymous · Answer

Do you know where to start?

anonymous · Answer

Derivative possibly?

anonymous · Answer

Yes!

anonymous · Answer

So, you'll need to take the first, and second derivative of that function.

anonymous · Answer

ok, so it would be something like y' = x - 1/x ?

anonymous · Answer

Uhh, close, it should be y' = 1 - 1/x

anonymous · Answer

Because the x turns into a 1.

anonymous · Answer

oh right!

anonymous · Answer

And now for the second derivative, we know that 1 will just go to 0, and you'll be left with the -1/x portion. 
Change it to $$-x ^{-1}$$
And now just follow the exponent rule for derivatives.

anonymous · Answer

so y'' would be 1/x^2

anonymous · Answer

$$y'=1-\frac{ 1 }{ x }$$$$y''=\frac{ 1 }{ x^2 }$$And now find where y equals 0 in each problem.

anonymous · Answer

But, hmm, the 1st one is easy, but the second one, actually can't equal 0 from what I see, so there might not be any inflection points.

anonymous · Answer

But the relative extrema would be at x = 1

anonymous · Answer

I don't quite understand how you find where y equals 0.

anonymous · Answer

Oh, well, if you plug in 1 for x in the y' equation, notice it's 1 - 1/1, or just 1 - 1 which equals 0.

anonymous · Answer

Ah, that makes sense.

anonymous · Answer

You just set the equation equal to 0 and solve for x if the equation is a bit trickier.

anonymous · Answer

Yeah, there doesn't appear to be any inflection points, if you were to look at a graph of 1/x^2, you'll notice it doesn't cross the x axis, meaning that it doesn't change from concave up to concave down, vice versa.

anonymous · Answer

so there is only one relative extrema then and no inflection points?

anonymous · Answer

Yeah, it appears so.

anonymous · Answer

ok, thank you very much, do you have time to help me with one more problem?

anonymous · Answer

Sure.

anonymous · Answer

Alrighty, it is: find an equation of the tangent line to the graph of f at the given point.
f(x) = 4 - x^2 - ln(1/2x + 1), (0,4)

anonymous · Answer

Ah, easy. Just find the derivative of that function, plug in 0 for x, and see what the slope is there.

anonymous · Answer

We're going to form the equation y = mx + b

anonymous · Answer

They give you the y intercept, which is 4, we just need to find m, or slope at x = 0

anonymous · Answer

ah ok, I think the derivative is -2x + 1/(2x+1)(2) ?

anonymous · Answer

Uhh, I actually have to go sorry!

anonymous · Answer

ok, thats fine. I should be able to get it from here, thanks so much!

anonymous · Answer

Okay, good luck!