Expert: Paul Klarreich - 9/25/2008

Question
Determine the point(s) at which the graph of f(x) = x/sqrrt(2x-1) has a horizontal tangent

Answer
Questioner:   Kristin
Category:  Calculus
Private:  No
 
Subject:  Horizontal tangents
Question:  Determine the point(s) at which the graph of f(x) = x/sqrt(2x-1) has a horizontal tangent
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Hi, again,

Use the vocabulary:

the graph has a horizontal tangent

MEANS

the slope of the tangent is zero,

WHICH MEANS

the derivative of the function is zero.

So you know what to do:  Find f'(x), set it to zero, solve.

For f(x) = x/sqrt(2x-1), use the quotient rule.  Do it like this:

Write:
       ()() - ()()
f'(x) = --------------
           ()^2

Fill in:

       (sqrt(2x-1))(1) - (x)(1/sqrt(2x-1))
f'(x) = -----------------------------------
             (sqrt(2x-1))^2

Do some simplifying:

       sqrt(2x-1) - x/sqrt(2x-1)
f'(x) = --------------------------
            (2x-1)

Multiply out sqrt(..)

       2x-1  - x
f'(x) = ----------------
        (2x-1)^3/2

        x - 1
f'(x) = -----------
       (2x-1)^3/2

Now set that (just the top, actually) to zero and get

x = 1.