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

Question

Ezra Adams · Accepted Answer

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).

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))#

Paisley Johnson · Answer

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)).# 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))#

Aurora Alvarado · Answer

undefined To find the points where the graph of the function y=2x(8-x)^.5 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function y=2x(8-x)^.5 using the product rule and chain rule.

The derivative is given by: dy/dx = 2(8-x)^.5 + 2x(0.5)(8-x)^(-0.5)(-1)

Setting the derivative equal to zero, we have: 2(8-x)^.5 + 2x(0.5)(8-x)^(-0.5)(-1) = 0

Simplifying the equation, we get: (8-x)^.5 - x(8-x)^(-0.5) = 0

To solve this equation, we can square both sides: (8-x) - x^2(8-x)^(-1) = 0

Expanding and rearranging, we have: 8 - x - 8x^2 + x^3 = 0

This equation can be factored as: (x-4)(x^2 - 4x + 2) = 0

Setting each factor equal to zero, we get two possible values for x: x = 4 and x = 2 ± √2

Therefore, the points where the graph of the function y=2x(8-x)^.5 has horizontal tangents are (4, 0), (2 + √2, 0), and (2 - √2, 0).