# finding the coordinantes for the tangent line.

• Jul 8th 2007, 10:05 AM
driver327
finding the coordinantes for the tangent line.
Find the coordinates of each point on the graph of the given function where the tangent line is horizontal.
I have this problem:
f(x)=(sqrt x)(x-3).
I used the product rule:
(sqrt x)(x-3)'+(sqrt x)'(x-3)
coming out (sqrt x)+x-3. I know this cannot be right.
What am I doing wrong with this?
The answer is suppose to be (1,-2).
• Jul 8th 2007, 10:21 AM
galactus
You should get $f'(x)=\frac{3(x-1)}{2\sqrt{x}}$

As you can see, what makes the numerator equal to 0?.
• Jul 8th 2007, 10:31 AM
driver327
Quote:

Originally Posted by galactus
You should get $f'(x)=\frac{3(x-1)}{2\sqrt{x}}$

As you can see, what makes the numerator equal to 0?.

From what you wrote, I understand how the numerator get to zero, But I am lost on how to get it to that particular form.
• Jul 8th 2007, 10:44 AM
red_dog
$f(x)=(x-3)\sqrt{x}$
$f'(x)=(x-3)'\sqrt{x}+(x-3)(\sqrt{x})'=\sqrt{x}+\frac{x-3}{2\sqrt{x}}=\frac{3x-3}{2\sqrt{x}}=\frac{3(x-1)}{2\sqrt{x}}$
• Jul 8th 2007, 02:04 PM
Soroban
Hello, driver!

Quote:

Find the coordinates of each point on the graph of the given function
where the tangent line is horizontal: . $f(x)\;=\;\sqrt{x}(x-3)$

We have: . $f(x)\;=\;x^{\frac{1}{2}}(x - 3) \;=\;x^{\frac{3}{2}} - 3x^{\frac{1}{2}}$

Then: . $f'(x)\;=\;\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}}\;=\;0\quad\Rightarrow\quad\frac{3}{2} \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)\:=\:0\quad\Rightarrow\qu ad \sqrt{x} - \frac{1}{\sqrt{x}} \;=\;0$

Multiply through by $\sqrt{x}\!:\;\;x - 1\:=\:0\quad\Rightarrow\quad x \,=\,1$

There is a horizontal tangent at $(1,\,-2)$