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Local Max and Min Notes » Calculus » Local Max and Min

The ability to find the slope of a function allows us to do a lot… This page is just the tip of that iceberg so to speak.

A question that the IB loves (and most math teachers), is to find the local maximum and minimum of a function. For example the function $f(x)=x^3+3x^2$. Given your GDC you can plot the graph and have the calculator find the max and mins. But clearly I wouldn't have written so much if that was the whole story. With a bit of soul searching you might realize that the slope of a function at a local max or min will be zero! In other words $f^\prime (x)=0$. For this example that means:

(1) \begin{equation} f'(x)=3x^2+6x=0 \end{equation}

This function can be easily factored to find the zeros (solutions):

(2) \begin{equation} f'(x)=(3x)(x+2)=0 \end{equation}

Thus the zeros or solutions are $x=0$ and $x=-2$. The solutions are the points where the slope of the original function $f(x)$ equals zero and thus they are the points where the local max and min occur. Check the graph below (the original function is in blue and the derivative is in red). Notice that the point where the derivative function crosses the x-axis is directly below the local maximum!

The method above fairly quickly found the local max and min, but it did not make a distinction between them… Sure we can look at our graph and clearly see which is which, but maybe we can use the calculus to go one step further.

Maximums - Notice that the slope of the function just before (to the left) of the maximum is positive and just after (to the right) the slope is negative. This will always be true for a maximum (try drawing a local maximum where this is not true). This also results in a derivative function that is positive before the maximum and negative after the maximum. This means that the slope of the derivative function is negative at the point where the original function has a maximum. Whoa. Might want to reread that.

If we are talking about the slope of a function that means one thing. Derivative! The derivative function is just like any other function so we can take the derivative of the derivative function, which is commonly referred to as the second derivative and is denoted by the notation:

(3) \begin{align} second \: \: \: derivative \rightarrow f''(x) = \frac {d^2 y}{dx^2} \end{align}

Find the second derivative (or third or fourth) of a function follows all of the same rules as finding the first derivative.

Minimums - The opposite of the maximum can be noticed. That is the slope of the original function before the minimum is negative. The slope of the original function after the minimum is positive. This results in the slope of the first derivative function being positive.

1st Derivative	2nd Derivative	Concavity
Maximum	$f'(x)=\frac{dy}{dx}=0$	$f''(x)=\frac{d^2y}{dx^2}<0$	Concave down
Minimum	$f'(x)=\frac{dy}{dx}=0$	$f''(x)=\frac{d^2y}{dx^2}>0$	Concave up

It can be easy to get sloppy and say that the second derivative is not needed and that the first derivative is all we need to find max and mins, this is often true, but there are a few cases which it is not true. For example in the function $g(x)= \frac{1}{3}x^3+2x+4x$. The derivative of this function is:

(4) \begin{equation} g'(x)=x^2+4x+4 \end{equation}

The zeros of the derivative function are repeated zeros at $x=-2$. In the graph below the original function is shown in blue and the derivative function is shown in red.

Notice that the (first) derivative is zero, but clearly $x=-2$ is not a maximum of the original function. The slope of the derivative function is neither positive or negative, but rather it is zero! Which means that at $x=-2$ that:

(5) \begin{equation} g''(x)=0 \end{equation}

Points of Inflexion (Inflection) - Are points where the original function changes concavity (up/down). In the graph above the point of inflexion is at $x=-2$. To the left of the inflexion point, the original (blue) function is concave down. To the right of the inflexion point, the original function is concave up. In general points of inflexion occur when the second derivative is zero. However in math lingo, this is "necessary but not sufficient."

An example of where this is not sufficient is the function $h(x)=x^4$ the first and second derivatives are:

(6) \begin{equation} h'(x)=4x^3 \end{equation} (7) \begin{equation} h''(x)=12x^2 \end{equation}

Graphing these three functions we have:

Notice that the second derivative function (shown in green) does in fact equal zero at $x=0$, but the original function is concave up on both sides of the origin (x=0). Therefore despite the fact that $h''(0)=0$ the point $(0,0)$ is not an inflexion point.

In order for a point to be an inflexion point two things must be true.

1. $h''(x)=0$
2. The second derivative must change signs (+/-). i.e. be positive (negative) to the left and be negative (positive) to the right.

Want to add to or make a comment on these notes? Do it below.

Show Comments Hide All Comments Unfold All Fold All Fold David (guest) 04 May 2013 14:18

For your last point do you not mean the first derivative must change signs? The second derivative cant if it is 0.

Reply Options Unfold by David (guest), 04 May 2013 14:18 Fold Arnav (guest) 07 May 2013 17:45

For equation (4), should g(x)=(1/3)x^3+2x^2+4x?

Reply Options Unfold by Arnav (guest), 07 May 2013 17:45 Add a New Comment Post preview: Close preview

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