SOLUTION: How do you determine the point (x,y) at which the graph of {{{ f(x) = (-8x)/sqrt(2x-1) }}} has a horizontal tangent? So far I get d/dx to be {{{(-8(2x-1)-8x)/((2x-1)^(3/2))}}} bu

Algebra ->  Test -> SOLUTION: How do you determine the point (x,y) at which the graph of {{{ f(x) = (-8x)/sqrt(2x-1) }}} has a horizontal tangent? So far I get d/dx to be {{{(-8(2x-1)-8x)/((2x-1)^(3/2))}}} bu      Log On


   

Question 1123335: How do you determine the point (x,y) at which the graph of +f%28x%29+=+%28-8x%29%2Fsqrt%282x-1%29+ has a horizontal tangent?
So far I get d/dx to be %28-8%282x-1%29-8x%29%2F%28%282x-1%29%5E%283%2F2%29%29 but what do you do to get a point value?
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Keep going.

df%2Fdx=%28-8x%2B8%29%2F%28%282x-1%29sqrt%282x-1%29%29

This derivative is 0 at x=1, and f%281%29=-8.