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# Find Stationary Points and Inflexions of a cubic polynomial

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The problem looks at a general cubic polynomial, and calculates the conditions needed for exactly two stationary points to exist. It also finds the inflexion point.

Consider the cubic polynomial (degree 3) given by, y=ax^3+bx^2+cx+d, where a is not equal to 0.

(a) Find the condition on the constants a,b,c so that this function has two stationary points.

(b) Find the x-values of the stationary points (and do not try to find the y-values).

(c) Find the coordinates of the point of inflexion,that is find the x and y values of the point of inflexion.

(d) Do a specific example with numbers substituted for a,b,c and where two stationary points occur. Draw a sketch of the cubic showing the y-intercept and the point of inflexion. You can use a calculator to get a rough idea of the y coordinates of the stationary points (should they turn out to be hard to find exactly).

##### Solution Summary

Calculus is used to find stationary points and inflexion point for the general case and a condition is deduced to ensure two stationary points exist.

An explicit example is done for a particular cubic polynomial, and a rough graph is sketched.

##### Solution Preview

The whole solution including a diagram are in the attached pdf file (and is easier to read).

The solution below is a text version.

Differentiating we have the two derivatives:

y' = 3ax^2+2bx+c
y'' = 6ax+2b

For stationary points we solve: y'=0

x=(-2b+- sqrt{4b^2-12ac} )/ 6a

x=(-b+-sqrt{b^2-3ac} )/ 3a}

There are three cases

(1) When b^2-3ac>0 there are two stationary points
...

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