Inflection Point of y=2√x-x: Derivative Analysis

[phillyolly]

Homework Statement

y=2√x-x

The Attempt at a Solution

First derivative:
-2x^-2-1
Second derivative:
4x^-3

4x^-3=0
No solutions?

 

[Mentallic]

\sqrt{x}=x^{1/2}

So y=2x^{1/2}-x

Then \frac{dy}{dx}=x^{-1/2}-1

Which isn't what you got.

 

[Mentallic]

But even so, if you correctly find the second derivative you'll see that it cannot equal 0, which means there is no inflection point. This means the graph of y=f(x) doesn't ever go from concave up to concave down or vice versa, it is always just one. Can you figure out whether it is always concave up or down without looking at the graph?

 

[phillyolly]

But even so, if you correctly find the second derivative you'll see that it cannot equal 0, which means there is no inflection point. This means the graph of y=f(x) doesn't ever go from concave up to concave down or vice versa, it is always just one. Can you figure out whether it is always concave up or down without looking at the graph?

So the first derivative is
(x^-1/2-1)
I take a second derivative to find an inflection point.
y''=-1/2x^-3/2 Hence, x=0...

What does it mean?...

 

[Bohrok]

y'' = -\frac{1}{2}x^{-\frac{3}{2}} = -\frac{1}{2x^{3/2}}

x = 0 isn't in the domain because you would have 0 in the denominator, but that's not really important. This function y'' is never 0, it will never touch the x-axis, so y does not have an inflection point.

 

[phillyolly]

y'' = -\frac{1}{2}x^{-\frac{3}{2}} = -\frac{1}{2x^{3/2}}

x = 0 isn't in the domain because you would have 0 in the denominator, but that's not really important. This function y'' is never 0, it will never touch the x-axis, so y does not have an inflection point.

Yes, I see I was wrong, I cannot have x=0 as a solution in this function.