Question

How do you find the points where the graph of the function `y=2x(8-x)^.5` has horizontal tangents?

Answer 1

You compute the first derivative, set that equal to zero, solve for the x value(s), and then use the function to give you the corresponding y value(s).

Explanation:

Compute the first derivative:

`y' = (16 - 3x)/sqrt(8 - x)`

Set that equal to zero:

`(16 - 3x)/sqrt(8 - x) = 0`

`x = 16/3`

`y(16/3) = 2(16/3)(8 - 16/3)^0.5`

The horizontal tangent is at `(16/3, (64/3)sqrt(2/3))`

Answer 2

The curve `graph{2x(8-x)^(0.5) [-46.65, 57.35, -8.66, 43.4]}` has a horozontal tangent at the point

`(16/3,64/3sqrt(2/3)).`

Explanation:

For the horizontal tgts., the slope of tgt., or, what is the same as to

say, `dy/dx`, must be `0`.

`dy/dx=0 rArr d/dx{2x(8-x)^(1/2)}=0`

`rArr 2{xd/dxsqrt(8-x)+sqrt(8-x)d/dx(x)}=0`.

`rArr x(1/(2sqrt(8-x)))(-1)+sqrt(8-x)=0`

`rArr sqrt(8-x)=x/(2sqrt(8-x))`

`rArr 2(8-x)=x`

`rArr x=16/3`

When, `x=16/3, y=2(16/3)sqrt(8-16/3)=32/3(2sqrt2)/sqrt3=64/3sqrt(2/3)`

It follows that, the curve `C : y=2x(8-x)^(0.5)` has a horozontal

tangent at the pt. `(16/3,64/3sqrt(2/3))`