Extreme Values

Extreme Values refer to either the maximum (the largest value of a function) or the minimum (the smallest value of a function). This lesson will only focus on Extreme Values pertaining to closed intervals.

Relative Values

The Relative Minimum is the point of a function where it changes direction from decreasing to increasing. The Relative Maximum is the point of a function where it changes direction from increasing to decreasing.

Relative Max and Min are used synonymously with Local Max and Min. These points only exist at the turning points (within the closed interval) instead of at the end points of the interval.


Absolute Values

The Absolute Maximum is the largest point of a function whereas the Absolute Minimum is the smallest point of a function. Unlike Relative Values, Absolute Values can either exist as the turning points or end points. If a turning point is either higher or lower than the respective endpoints, then it will be classified as an Absolute Value.


Graph showing the Abosolute Maximum and Minimums of a function.
Graph showing the Absolute and Relative Extrema of a function.

NOTE: A continuous function with a closed interval will ALWAYS have an absolute max and min, and may OCCASIONALLY have a relative max and/or min.


Critical Numbers

Critical Numbers refer to extreme values of a function. These can be determined when either:

  • \(f'(x) = 0\)
  • \(f'(x) = \text{undefined}\)

In summary, the critical points can be found by identifying the \(x\) values that make the function's derivative either \(0\) or undefined.

Since the derivative can be used to represent the slope of the original function, if the derivative is \(0\) at a point, that means the original function has a horizontal tangent at that point.


Example

Find the absolute maximum and minimum of the function \(f(x) = x^3 -12x-3\) on the interval \(-3\leq t \leq 4\)

We first need to determine the derivative of the original function:

\(f'(x) = 3x^2 -12\)

We can now set the function's derivative equal to \(0\) and factor it to determine its critical points:

\(0 = 3x^2 -12\)

\(0 = 3(x^2 -4)\)

\(0 = 3(x-2)(x+2)\)

We can determine that the critical points are \(x = -2\) and \(x = 2\). Since they are within the interval, we can use these values to determine the absolute value.

We can then substitute the local extreme values for \(x\) in the original function to determine their respective \(y\)-coordinates:

\(f(-3) = (-3)^3 -12(-3)-3\)

\(f(-3) = 6\)

\(f(-2) = (-2)^3 -12(-2)-3\)

\(\textcolor{red}{f(-2) = 13}\)

\(f(2) = (2)^3 -12(2)-3\)

\(\textcolor{blue}{f(2) = -19}\)

\(f(4) = (4)^3 -12(4)-3\)

\(\textcolor{red}{f(4) = 13}\)

Therefore, we can determine that the absolute minimum is located at \((2,-19)\). We can also determine that absolute maximum is located at both \((-2,13)\) and \((4,13)\).


A section of the roller coaster is in the shape of \(f(x) = -x^3-2x^2 +x+15\) where \(-2\leq x \leq 2\).

  1. Find all local extreme values
  2. Is the highest point of this section of the ride at the beginning, the end, or neither?

Try these questions: