How can I solve this?

Question

Consider the curve defined by y^2=x^3−3x+3 for x>0. At what value of x does the curve have a horizontal tangent?

a) √3 / 3

b) 1

c) √3

d) There is no such value

Thank you

Answer 1

The solution suggested above isn't completely correct. If you plot the function suggested we have horizontal tangents at the turning points x = -1, x = +1

By implicit differentiation

2y dy/dx = 3x² - 3

dy/dx = (3x²-3) / 2y

dy/dx is slope of tangent line, so when tangent line is horizontal dy/dx = 0

So assuming y is non zero, dy/dx will be zero when 3x² - 3 = 0, x² = 1, x = +1 or x = -1

Answer 2

Hi Amber L.,

I tried setting this up by implicit differentiation and by brute force, but seem to have made errors! But you can graph the function y = (3x^3-3x+3)^0.5 and observe horizontal tangents at ± 1/√3 .

-- Cheers, -- Mr. d.