What is the general way of calculating maxima and minima of a given cubic function?

Question

What is the general way of calculating maxima and minima of a given cubic function (ax^3 + bx^2 + cx + d) with variables (including a varible leading co-effienient) with a closed interval (α ≤ x ≤ β). I am struggling to make a general set of steps to follow; here is what I've got so far:

1. Differentiate the given cubic function and factorize to determine the critical values or relative extremes
2. Draw up a variation table with x, f'(x) and f(x) as well as α and β
3. Compare f(x), f'(x) to verify the shape of the graph and identify maxima and minima and the co-ordinates

But this is were I start getting confused; the cannot correctly give the inequality of the variable, particularly if the leading co-effieient is a variable as it changes the shape of the graph, where the maxima and minima changes.

Answer 1

You need to consider extremes and turning points.

For turning points look for $f'(x)=0$, and here if $f(x)=ax^3 + bx^2 + cx + d$ then $f'(x)=3ax^2+2bx+c$. So

• If $b^2 \gt 3ac$ then let $x_0 = \frac{-b+\sqrt{b^2-3ac}}{3a}$ and $x_1 = \frac{-b-\sqrt{b^2-3ac}}{3a}$
• If $b^2 = 3ac$ then let $x_0 = -b$

Now consider the values of

• $f(\alpha)$
• $f(\beta)$
• $f(x_0)$ if you have found it earlier and $x_0 \in [\alpha,\beta]$
• $f(x_1)$ if you have found it earlier and $x_1 \in [\alpha,\beta]$

The largest of these will be the maximum and the smallest will be the minimum

Answer 2

I'm not clear what you mean by "cannot correctly give the inequality of the variable." The leading coefficient of the cubic won't affect the general strategy. Perhaps an example will help clarify things:

Suppose we want to find the maximum of $f(x) = 2x^3 - 3x^2 - 12x + 5$ on $[-2,1]$.

1. $f'(x) = 6x^2 - 6x - 12 = 6(x^2-x-2) = 6(x-2)(x+1).$ Thus, $f'(x) = 0$ when $x = 2,-1$. So the critical values are $x = \pm 1, \pm 2$.

2. I will leave writing the table to you, but note that $f'(x)$ is an upward-facing parabola with roots $x = 2$ and $x = -1$. By sketching a graph of $f'(x)$, you will see that $f'$ changes from positive to negative at $x = -1$, and from negative to positive when $x = 2$. Thus, $f(-1) = 9$ is a local max (and $f(2) = -18$ is a local min). Now checking the endpoints of the interval: $f(-2) = -2$ and $f(1) = -11$. Thus, the maximum value of $f(x)$ on $[-2,1]$ is 9.

Does this make more sense now?