We refer to the maximum and minimum values as ______ ______ or ______ (singular: ______) and to the process of finding them as _________. Sometimes, we are interested in finding the min or max for x in a particular interval I, rather than on the entire domain of f
- extreme values or extrema
- singular: extremum
Define: Extreme values on an Interval
Let f be a function on an interval I and let a ∈ I. We say that f(a) is the
- Absolute minimum of f on I if f(a) ≤ f(x) for all x ∈ I
- Absolute maximum of f on I if f(a) ≥ f(x) for all x ∈ I
Does every function have a minimum or maximum value? If not, state an example
Clearly not, as we see by taking f(x) = x. Indeed, f(x) increases without bound as x → ∞ and decreases without bound as as x → -∞
In fact, extreme values do not always exist even if we restrict ourselves to an interval I. Name two scenarios where things can go wrong
- Discontinuity: (Fig 2A pg 175) shows a discontinuous function with no maximum value. The values of f(x) get arbitrarily close to 3 from below, but 3 is not the maximum value because f(x) never actually takes on the value 3.
- Open interval: In (Fig 2B), g(x) is defined on the open interval (a,b). It has no max because it tends to ∞ on the right, and it has no min because it tends to 10 on the left without ever reaching this value.
Explain the Existence of Extrema on a Closed Interval Theorem
A continuous function f on a closed (bounded) interval I = [a,b] takes on both a minimum and a maximum value on I.
Define both cases of Local Extrema:
We say that f(c) is a
- Local minimum occurring at x = c if f(c) is the minimum value of f on some open interval (in the domain of f) containing c
- Local maximum occurring at x = c if f(c) is the maximum value of f on some open interval (in the domain of f) containing c
A local max occurs at x = c if (c, f(c)) is the highest point on the graph within some small box [Fig 3(A) pg 176]. Thus, f(c) is greater than or equal to all other nearby values, but it does not have to be the ______ ______ value of f.
Local minima are similar. Fig 3B (pg 176) shows the difference between local and absolute extrema: f(a) is the absolute max on [a,b] but is not a local max. (Why?)
because f(x) takes on larger values to the left of x=a
What is the crucial observation when finding local extrema? (explain). What if f'(c) is not differentiable?
- The crucial observation is that the tangent line at a local min or max is horizontal [fig. 4A pg 176]. In other words, if f(c) is a local min or max, then f'(c) = 0.
- Then the tangent line may not exist
Define Critical Points
A number c in the domain of f is called a critical point if either f'(c) = 0 or f'(c) does not exist
Fermat's Theorem on Local Extrema
If f(c) is a local min or max, then c is a critical point of f
Extreme Values on a Closed Interval Theorem
Assume that f is continuous on [a,b] and let f(c) be the minimum or maximum value on [a,b]. Then c is either a critical point or one of the endpoints a or b
As an application of our optimization methods, we prove Rolle's Theorem: If f is differentiable between two points a and b, then somewhere between these two points, the derivative is ____. Graphically, what happens if the secant line between x = a and x = b is horizontal?
- At least one tangent line between a and b is also horizontal
Formally state Rolle's Theorem
Assume that f is continuous on [a,b] and differentiable on (a,b). If f(a) = f(b), then there exists a number c between a and b such that f'(c) = 0
Rolle's Theorem Proof: Since f is continuous and [a,b] is closed, f has a min and a max in [a,b]. Where do they occur? Specifically what happens if either the min or max occurs at a point c in the open interval (a,b)? What happens if the min and max occur at the endpoints and f(a) = f(b)?
- If the min or max occurs at a point c in the open interval, then f(c) is a local extreme value and f'(c) = 0 by Fermat's Theorem.
- If the min and max occur at the endpoints and f(a) = f(b), then the min and max coincide and f is a constant function with zero derivative. Then f'(c) = 0 for all c in (a,b)
Verify Rolle's Theorem for:
f(x) = x4 - x2 on [-2,2]
Hint: Illustrating Rolle's Theorem
- The hypotheses of Rolle's Theorem are satisfied because f is differentiable (and therefore continuous) everywhere, and f(2) = f(-2):
- f(2) = 24 -22 = 12, f(-2) = (-2)4 - (-2)2 = 12
- We must verify that f'(c) = 0 has a solution in (-2,2) so we solve f'(x) = 4x3 -2x = 2x(2x2 - 1) = 0.
- The solution are c = 0 and c = ±1/√2 ~ ±0.707. They all lie in (-2,2), so Rolle's Theorem is satisfied with three values of c
Show that f(x) = x3 + 9x - 4 has precisely one real root
Hint: Use Rolle's Theorem
- First, we note that f(0) = -4 is negative and f(1) = 6 is positive.
- By the Intermediate Value Theorem (Section 2.8), f has at least one root a in [0,1].
- If f had a second root b, then f(a) = f(b) = 0 and Rolle's Theorem would imply that f'(c) = 0 for some c ∈ (a,b).
- This is not possible because f'(x) = 3x2 + 9 ≥ 9, so f'(c) = 0 has no solutions.
- We can conclude that a is the only real root of f (fig. 14)