# stationary point

• November 16th 2008, 01:24 AM
Derivative of Man
stationary point
im having trouble understanding my textbook.

Question: Use Calculus to determine any stationary points on the curve
$y=2x+(6/x)$.

i found that x= √3.
the book then explians that
y=2√3+(6/√3)(√3/√3)

where does the extra √3/√3 come from?
• November 16th 2008, 01:45 AM
mr fantastic
Quote:

Originally Posted by Derivative of Man
im having trouble understanding my textbook.

Question: Use Calculus to determine any stationary points on the curve
$y=2x+(6/x)$.

i found that x= √3.
the book then explians that
y=2√3+(6/√3)(√3/√3)

where does the extra √3/√3 come from?

They are rationalising the denominator:

$y = 2 \sqrt{3} + \frac{6}{\sqrt{3}} = 2 \sqrt{3} + \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = 2 \sqrt{3} + \frac{6 \sqrt{3}}{3} = 2 \sqrt{3} + 2 \sqrt{3} = 4 \sqrt{3}$.