Solve for m & n for a cubic using derivatives (stationary points)

Question

If f(x) = x^3+mx^2+nx and has stationary points at x=2 & x= (-4/3) find coefficients m & n. Please help

Answer 1

A stationary point is a point where the derivative of a function is zero.  Hence it will be a local maximum or minimum.

 

     f(x) = x³ + mx² + nx

 

Take the first derivative wrt x and set it to zero:

 

     f '(x) = 3x² + 2mx + n

     0 = 3x² + 2mx + n

 

Use the quadratic formula to solve for x:

 

     x = (-m±√(m²-3n))/3

 

Now we are given that x = -4/3 and 2 at the stationary points.

 

     2 = (-m+√(m²-3n))/3

     6 = -m+√(m²-3n)

 

     -4/3 = (-m-√(m²-3n))/3

     -4 = -m-√(m²-3n)

 

To find m, add the two equations together:

     6 + (-4) = -m+√(m²-3n) + -m-√(m²-3n)

     2 = -2m

     -1 = m

 

To find n, substitute m=-1 into either quadratic solution:

     -4 = -m-√(m²-3n)

     -4 = -(-1) - √((-1)²-3n)

     -4 = 1 - √(1-3n)

     -5 = -√(1-3n)

     25 = 1 - 3n

     24 = -3n

     -8 = n

 

f(x) = x³ - x² - 8x