At what point?

Question

If f(x)=2x^3-6x, at what point on the interval 0≤x≤sqrt(3), if any, is the tangent to the curve parallel to the secant line on that interval? 

 

Answer: 1

 

Please show all your work. 

Answer 1

f(x) = 2x³ - 6x

 

The slope of the secant line on the interval 0 ≤ x ≤ sqrt(3) is:

 

[ f(x=sqrt(3)) - f(x=0) ] / [ sqrt(3) - 0)] = 0 when you plug the values x = sqrt3 and x = 0 into the slope equation.  Hence the slope of the secant line is 0.

 

The Mean Value Theorem states that there must be at least one point between 0 and sqrt 3 where the tangent line has the same slope as the secant line.  The slope of the tangent line is given by:

 

df(x)/dx = 6x² - 6

 

Solve for 6x² - 6 = 0

 

x = 1

 

Note that x = -1 is also a solution but it does not lie within the specified interval.