A local minimum or local maximum of a function is an extreme value. A local extreme value may exist within an arbitrary open interval in the domain of a function. On a graph, a local extremum appears as a hill or valley. Local extrema are defined as follows:
A function f(x) has a local maximum value at a point p if and only if f(p) ≥ f(x) for all values of x in an open interval containing p.
A function f(x) has a local minimum value at a point p if and only if f(c) ≤ f(x) for all values of x in an open interval containing p.
Critical points
The graph of a polynomial of degree n has at most n – 1 extreme values. These extreme values occur only at critical points where the slope of a line tangent to the graph is zero. Calculus, which is beyond the scope of this text, allows us to derive equations that give the instantaneous slope of a function f(x). These equations are shown below for some named polynomials. The critical points may be found by finding the zeroes of the corresponding functions g(x). Note that not all critical points are extrema.
quadratic
f(x) = ax2 + bx + c g(x) = 2ax + b
cubic
f(x) = ax3 + bx2 + cx + d g(x) = 3ax2 + 2bx + c
quartic
f(x) = ax4 + bx3 + cx2 + dx + e g(x) = 4ax3 + 3bx2 + 2cx + d
quintic
f(x) = ax5 + bx4 + cx3 + dx2 + ex + f g(x) = 5ax4 + 4bx3 + 3cx2 + 2dx + e
Curvature
If, at the location of the critical point, the graph is either concave up or concave down, the critical point is a local extreme point. A region of a graph that is concave up has positive curvature; if concave down, the region has negative curvature. Calculus also allows us to derive equations that give the curvature of a function f(x). These equations are shown below for some named polynomials.
quadratic
f(x) = ax2 + bx + c h(x) = 2a
cubic
f(x) = ax3 + bx2 + cx + d h(x) = 6ax + 2b
quartic
f(x) = ax4 + bx3 + cx2 + dx + e h(x) = 12ax2 + 6bx + 2c
quintic
f(x) = ax5 + bx4 + cx3 + dx2 + ex + f h(x) = 20ax3 + 12bx2 + 6cx + 2d
If the curvature of a graph at a critical point is zero, we must use the definition of a local extreme value to determine whether the point is a maximum or a minimum. For example, consider the quartic function:
The critical points are the solution to:f(x) = 2x4 + 1, then g(x) = 8x3, and h(x) = 24x2
Which is: x = 0. Now,0 = 8x3
The curvature is zero, so this test fails to tell us whether the critical point is a maximum or minimum. Now consider points in the open interval (-c, +c).h(0) = 24(0)2 = 0
That is, for any choice of a small distance, c, from the critical point p = 0, there exists an open interval around x = 0 such that f(0) ≤ f(x). That is, f(0) is a local minimum.f(0) = 1 f(+c) = 2c4 + 1 f(-c) = 2c4 + 1