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The intermediate value theorem, which states that if a continuous function on a closed and bounded interval has a value l between its endpoints, then there exists a number c in the interval where the function takes on the value l. The document also discusses the concept of extreme values of a function, including local maximum and minimum arguments, and provides examples and theorems to illustrate these concepts.
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Figure 1. Intermediate Value Theorem
Figure 2. Intermediate Value Theorem
Definition 1. (Critical Point):
Let f (x) be a differentiable function defined on an open interval I = (a, b), with a < b. Let x = x 0 be a real number which belongs to the interval I = (a, b).
We say that x 0 is a critical point of the function f (x) if the derivative of f is zero at x 0 , i.e.,
(2.1) f โฒ(x 0 ) = 0.
Example 3. Let f (x) = x^3 โ x. Find the critical points of f (x) on (โโ, โ).
To find the critical points of f (x), the first step is to evaluate the derivative f โฒ(x). f โฒ(x) = 3x^2 โ 1.
The second step is to solve for x in the equation
f (x) = 0.
So the equation we need to solve is
3 x^2 โ 1 = 0.
We find
x =
or
x =
Our conclusion is that 13 , โ 31 are the only critical points of f (x) = x^3 โ x.
Remark 2: Note that critical points are points x 0 , where, the tangent lines (to the graph of f (x) at (x 0 , f (x 0 )) are horizontal. For example, look at the graph of the function f (x) = (x โ 1)^2 + 2 below in figure 3.
Example 4. Let f (x) = x^3 โ x^2 โ x โ 1. Find the critical points of f (x) on (โโ, โ).
Again, to find the critical points of f (x), the first step is to evaluate the derivative f โฒ(x). f โฒ(x) = 3x^2 โ 2 x โ 1.
The second step is to solve for x in the equation
f (x) = 0.
So the equation we need to solve is
3 x^2 โ 2 x โ 1 = 0.
Note that 3 x^2 โ 2 x โ 1 = (3x + 1)(x โ 1)
, and so from the equation 3 x^2 โ 2 x โ 1 = 0
we find that
x =
or x = 1.
Our conclusion is that x = 1, and x = โ 31 are the only critical points of f (x) = x^3 โ 2 x โ x โ 1.
Definition 2. (Local Extremum): Let f (x) be a continuous function defined on an open interval I = (a, b), with a < b. Let x = x 0 be a real number which belongs to the interval I = (a, b).
We say that x 0 is a local maximum argument for the function f (x) if there exists an open interval J = (ฮฑ, ฮฒ) centered at x 0 such that, for every point t belonging to J we have
(2.2) f (t) โค f (x 0 ).
Similarly, We say that x 0 is a local minimum argument for the function f (x) if there exists an open interval J = (ฮฑ, ฮฒ) centered at x 0 such that, for every point t belonging to J we have
(2.3) f (t) โฅ f (x 0 ).
Together, local maximum and local minimum arguments are called local extremum or local extremes.
Figure 4. Local Extremum
So the equation we need to solve is
6 x^2 + 6x โ 12 = 0.
Note that
6 x^2 + 6x โ 12 = 6(x + 2)(x โ 1),
and so from the equation
6 x^2 + 6x โ 12 = 0
we find that
6(x + 2)(x โ 1) = 0.
Thus,
x = โ 2 ,
or
x = 1.
Our conclusion is that x = 1, and x = โ 2 are the only critical points of f (x).
The second derivative of f (x) is
f โฒโฒ(x) = 12x + 6.
We note that
f โฒโฒ(1) = 18 > 0 ,
and the second derivative test implies that x = 1 is a local minimum argument. Also, We note that
f โฒโฒ(โ2) = โ 18 < 0 ,
and the second derivative test implies that x = 1 is a local maximum argument.
Theorem 3. (Extreme Value Theorem) Let I be the closed interval [a, b]. Let f (x) be a continuous function on I. Then f (x) has a maximum and a minimum in I. In other words, there exists a point x = ฮฑ in I such that for all x in I,
(2.4) f (x) โค f (ฮฑ).
and there exists a point x = ฮฒ in I such that for all x in I,
(2.5) f (x) โฅ f (ฮฒ).
Moreover, ฮฑ is either one of the endpoints a, b of the given interval [a, b], or ฮฑ is a critical point or a sharp point. Also, ฮฒ is either one of the endpoints a, b of the given interval [a, b], or ฮฒ is a critical point or a sharp point.
Example 7. Example: f (x) = 2x^3 โ 3 x^2 โ 12 x โ 7. Find the maximum and minimum values taken by f (x) on the closed and bounded interval [โ 3 , 0].
The first step is to find the critical points of f (x) in the open interval (โ 3 , 0).
From our last example, we found that x = 1, and x = โ 2 are the only critical points of f (x). Note x = 1 is not in the open interval (โ 3 , 0).
The second step is to evaluate f (x) at the critical points in (โ 3 , 0), and at the end points of the closed interval [โ 3 , 0].
x f (x) โ 2 13 โ 3 2 0 โ 7
Comparing these values we conclude that the absolute maximum of f (x) on the interval [โ 3 , 0] is f (โ2) = 13, and the absolute minimum of f (x) on the interval [โ 3 , 0] is f (0) = โ 7.
