Thread: x values where tangent is 0

1. x values where tangent is 0

Hi!

I am not sure how to go about the following.

Let $f(x) = \sqrt{x^2-x}$.

Find any $x$ values where the slope of the line tangent to the graph of $f(x)$ would be equal to $0$.

Would I simply find the roots: x = 0; x = 1?
Those are the x-values where the tangent is 0.

2. Re: x values where tangent is 0

Originally Posted by Unreal
Hi!

I am not sure how to go about the following.

Let $f(x) = \sqrt{x^2-x}$.

Find any $x$ values where the slope of the line tangent to the graph of $f(x)$ would be equal to $0$.

Would I simply find the roots: x = 0; x = 1?
Those are the x-values where the tangent is 0.
What is the condition for the slope to be 0? What do you need to do with f(x)? The derivative, of course. You might find this rewrite helps:
$f(x) = \sqrt{x^2 - x} = (x^2 - x)^{1/2}$

Can you finish it from here?

-Dan

3. Re: x values where tangent is 0

$f^\prime(x) = \frac{2x-1}{2\sqrt{x^2-x}}$

Check if numerator is equal to zero (zero signifies asymptote):
$2x-1 = 0\ ;\ x = \frac12$

There are no points where the slope of the line tangent to the graph f(x) would be equal to 0.

4. Re: x values where tangent is 0

I think you need to say a bit more than that! You say that the numerator of the derivative is 0 at x= 1/2 What then prevents you from saying that the derivative is 0 at x= 1/2?
(You are correct. I'm not saying you aren't, just that you need to say why.)