When using functions to analyze situations, we are often interested in the maximum and minimum values that a function takes on. The maximum o r minimum value the function assumes. These are often called extreme values. We would like to know the extreme values and where, in the domain, they occur.
There are two views on extreme values.
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The global maximum value is the greatest value of the function.
\[ f(c) \text { is a global maximum if } f(x) \leq f(c) \text { for all } x \text { in the domain of } f \] -
The global minimum value is the least value of the function.
\[ f(c) \text { is a global minimum if } f(c) \leq f(x) \text { for all } x \text { in the domain of } f \]
A function has at most, one maximum value and one minimum value, but they can occur at multiple domain numbers.
Global extrema are also called absolute extrema.
- A global maximum may occur at multiple domain numbers.
- A function can also not have a maximum value.
- A global minimum may occur at multiple domain numbers.
- A function can also not have a minimum value.
In contrast to a single maximum or minimum value, a function may also have values which are the greatest value in their own little neighborhood of the domain, but not neccessarily the greatest overall value. We see these visually encoded as tops of hills and bottom of valleys on the graph. A function may have many of these local or relative extrema.
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\(f(c)\) is a local maximum value of the function \(f\), if there exists an \(\epsilon > 0\) such that \(f(x) \leq f(c)\) for all domain numbers, \(x\), within a distance of \(\epsilon \) from \(c\).
\[ f(c) \text { is a local maximum if } f(x) \leq f(c) \text { for all } \{ x \in domain \, | \, x \in (c - \epsilon , c + \epsilon ) \} \text { for some } \epsilon > 0 \] -
\(f(c)\) is a local minimum value of the function \(f\), if there exists an \(\epsilon > 0\) such that \(f(c) \leq f(x)\) for all domain numbers, \(x\), within a distance of \(\epsilon \) from \(c\).
\[ f(c) \text { is a local minimum if } f(c) \leq f(x) \text { for all } \{ x \in domain \, | \, x \in (c - \epsilon , c + \epsilon ) \} \text { for some } \epsilon > 0 \]
Local extrema are also called relative extrema.
Note: By default, all global extrema are automatically local extrema.
Let \(g(k)\) be a function. The graph of \(y= g(k)\) is displayed below.
![[Picture]](optimization-1bdc82fd211deda1501b0630f91b0ea9.svg)
- \(g\) has no global maximum.
- The global minimum of \(g\) is \(-3.3\), which occurs at \(-8\).
- \(g\) has a local minimum of \(-3.3\), which occurs at \(-8\) and a local minimum of \(-2.3\), which occurs at \(0.9\).
- \(g\) has a local maximum of \(2.4\), which occurs at \(-4.9\)
Let’s run through the idea of a local extrema for the previous example.
\(\blacktriangleright \) \(g\) has a local minimum of \(-2.3\), which occurs at \(0.9\).
Let \(\epsilon = 0.3\). Then, \(\epsilon > 0\).
\(-2.3 = g(0.9) \leq g(k)\) for all \(k\) in the domain within a distance of \(0.3\) from \(0.9\), which would be the interval
\(\left (\answer {0.6}, \answer {1.2}\right )\).
\(\blacktriangleright \) \(g\) has a local maximum of \(2.4\), which occurs at \(-4.9\).
Let \(\epsilon = 0.4\). Then, \(\epsilon > 0\).
\(2.4 = g(k) \leq g(-4.9)\) for all \(k\) in the domain within a distance of \(0.4\) from \(-4.9\), which would be the interval
\(\left (\answer {-5.3}, \answer {-4.5}\right )\).
\(\blacktriangleright \) \(g\) has a local minimum of \(-3.3\), which occurs at \(-8\).
Let \(\epsilon = 0.5\). Then, \(\epsilon > 0\).
\(-3.3 = g(-8) \leq g(k)\) for all \(k\) in the domain within a distance of \(0.5\) from \(-8\), which would be the interval \(\left [\answer {-8.5}, \answer {-7.5}\right )\).
Note: As the previous example illustrates, global extrema are also local extrema.
Note: We picked \(0.3\), \(0.4\), and \(0.5\) for \(\epsilon \), but we could have selected any small, positive numbers
that work. In fact, once one positive number works, then any lesser positive number
will also work.
Let \(B(w)\) be a function. The graph of \(y = B(w)\) is displayed below.
![[Picture]](optimization-f717b62c510fbce3f96218a5a601eb3d.svg)
- \(B\) has no global maximum.
- The global minimum of \(B\) is \(-2.3\), which occurs at \(1.4\).
- \(B\) has a local minimum of \(-2.3\), which occurs at \(1.4\) and a local minimum of \(-0.7\), which occurs at \(-5.9\).
- \(B\) has a local maximum of \(0.25\), which occurs at \(-3\)
Let \(0 < \epsilon = 0.6\), then \(-2.3 = B(1.4) \leq B(w)\) for all \(w\) in the domain within a distance of \(\answer {0.6}\) from \(\answer {1.4}\), which would be the interval \((0.8, 2)\).
Let \(0 < \epsilon = 0.25\), then \(-0.7 = B(-5.9) \leq B(w)\) for all \(w\) in the domain within a distance of \(\answer {0.25}\) from \(\answer {-5.9}\), which would be the interval \((-6.15, -5.65)\).
Let \(0 < \epsilon = 0.5\), then \(0.25 = B(w) \leq B(-3)\) for all \(w\) in the domain within a distance of \(\answer {0.5}\) from \(\answer {-3}\), which would be the interval \((-3.5, -2.5)\).
Let \(T(p)\) be a function. The graph of \(y= T(p)\) is displayed below.
![[Picture]](optimization-8d5610d0958efabbd8ab6560f4368965.svg)
- \(T\) has no global maximum.
- \(T\) has no global minimum.
- \(T\) has a local minimum of \(-2\), which occurs at \(-8\).
- \(T\) has a local minimum of \(\answer {-6}\), which occurs at \(3\).
- \(T\) has a local maximum of \(\answer {6}\), which occurs at \(-1\).
It appears that \(8\) would have been the global maximum of \(T\) occurring at \(-8\), however, the point \((-8, 8)\) is missing from the graph and \(8\) is not in the range.
The real numbers have a special property which makes Calculus work.
We can illustrate this property with the number \(8\).
There is no real number “just below \(8\)”.
If you choose any real number, \(r\), below \(8\) to be the possible candidate for the previous number, then there is the number \(\frac {8+r}{2}\).
You cannot identify a specific number as the “previous” number, so there isn’t one.
This is the idea behind space and open intervals.
Let \(0 < \epsilon = \answer {0.5}\). \(T(-8) \leq T(p)\) for all \(p\) in the domain within a distance of \(0.5\) from \(-8\), which would be the interval \([-8, -7.5)\).
Let \(0 < \epsilon = \answer {0.4}\). \(T(3) \leq T(p)\) for all \(p\) in the domain within a distance of \(0.4\) from \(3\), which would be the interval \((3.4, 2.6)\).
Let \(0 < \epsilon = \answer {0.3}\). \(T(p) \leq T(-1)\) for all \(p\) in the domain within a distance of \(0.3\) from \(-1\), which would be the interval \((-1.3, -0.7)\).
Inside the natural numbers there is the idea of “next”. Each natural number has a
next number.
However, the real numbers does not have a “next” property. There is no “next” number for any real number.
Let \(r\) be a real number.
Supposed \(N\) is the “next” real number. Then we have a problem, because the real
number \(\frac {r+N}{2}\) is between \(r\) and \(N\). Therefore, \(N\) cannot be the “next” real number. There is no
“next” real number.
A consequence of this is that each open interval contains an infinite number of
numbers.
Note: This is not true for closed intervals. For example, the closed interval \([3, 3]\) contains only one number. This is one reason we talk about open intervals.
\(\blacktriangleright \) Thinking Ahead
Linear functions have the same rate of change over any interval.
That is not quite true. They have a constant rate of change over an interval of the
form \([a, b]\), where \(a \ne b\).
As a mental exercise, if a linear funciton has the exact same rate of change over any
interval \([a, b]\), then doesn’t it feel like the linear function should have the same rate of
change over the interval \([a, a]\)?
Just thinking.
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more examples can be found by following this link
More Examples of Visual Behavior