Minimum and maximum values

Definitions of absolute and local minima/maxima

  • In this section, we see how to employ differentiation to help find minimum and maximum values.
    • Definition. Let \(f\) be a function with domain \(D\), and let \(c\) be a point in \(D\).
    • Then \(f(c)\) is the absolute minimum value of \(f\) on \(D\), if \(f(c)\le f(x)\) for all \(x\) in \(D\).
    • \(f(c)\) is the absolute maximum value of \(f\) on \(D\), if \(f(c)\ge f(x)\) for all \(x\) in \(D\).
    • \(f(c)\) is a local minimum value, if \(f(c)\le f(x)\), when \(x\) is near \(c\).
    • \(f(c)\) is a local maximum value, if \(f(c)\ge f(x)\), when \(x\) is near \(c\).
    • "\(x\) is near \(c\)" means: the property holds for \(x\) in some open interval containing \(c\).
    • The maximum and minimum values of \(f\) are also called extreme values.
    • The absolute extreme values of \(f\) are also called global extreme values.
  • Examples.
    • The function \(f(x)=x^2\) has \(f(0)=0\) as a global minimum value, but it does not even have a local maximum value.
    • The function \(f(x)=sin(x)\) has \(f(\frac{\pi}{2})=1\) as a global maximum value, and \(f(-\frac{\pi}{2})=-1\) as a global minimum value. Note that the same global extreme value can be taken up more than once. Here, we have \(f(\frac{\pi}{2}+2k\pi)=1\) for any integer \(k\).
    • We'll see soon how to verify that the function \(f(x)=(x+1)x(x-1)=x^3-x\) has \(f(-3^{-1/2})=-3^{-3/2}+3^{-1/2}\) as a local maximum value, and \(f(3^{-1/2})=3^{-3/2}-3^{-1/2}\) as a local minimum value. It does not have no global extreme values.

The extreme value theorem

  • The extreme value theorem.
    • Let \(f\) be a function, which is continuous on its domain a closed interval \(D=[a,b]\). Then \(f\) attains an absolute maximum value \(f(c)\), and an absolute minimum value \(f(d)\) at some numbers \(c,d\) in \(D\).
    • Sketch of proof. Let \(\mathrm{Im}(f)\) be the collection of the values \(f(a)\) for \(a\) in \(D\). It can be pictured as the projection of the graph of \(f\) to the \(y\)-axis. It is called the image of \(f\).
    • Since \(D\) is a closed interval, and \(f\) is continuous, its image \(\mathrm{Im}(f)\) is also a closed interval \([\alpha,\beta]\).
    • Then by construction \(\alpha=f(c)\) is a global minimum value, and \(\beta=f(d)\) is a global maximum value. QED (End of proof)
    • This is only a sketch, because it's not easy to show that the image \(\mathrm{Im}(f)\) is a closed interval.
    • Questions like that are answered in the field of Mathematics called Topology. It's a very interesting topic, I suggest learning about it if you have the chance!

Fermat's theorem and critical numbers

  • Fermat's theorem.
    • Let \(f\) be a function. Suppose that \(f(c)\) is a local extreme value of \(f\), and that \(f'(c)\) exists. Then we have \(f'(c)=0\).
    • Here the proof is easier, you can check it out in the textbook if you're interested.
  • Examples.
    • Let \(f(x)=x^2\). Then \(f'(x=0)=(2x)(x=0)=0\), and \(f(x=0)\) is a global minimum value.
    • Let \(f(x)=x^3-x\). Then \(f'(x)=3x^2-1\), the zeroes of which are \(x=\pm3^{-1/2}\). These are local extreme values.
    • Let \(f(x)=x^3\). Then \(f'(x=0)=(3x^2)(x=0)=0\), but \(f(x=0)\) is not a local extreme value.
    • Let \(f(x)=|x|\). Then \(|0|\) is a global minimum value, but \(f'(x=0)\) does not exist.
  • Summary of examples
    • It is possible that \(f'(c)=0\) even though \(f(c)\) is not a local extreme value
    • It is also possible that \(f(c)\) is a local extreme value, even though \(f'(c)\) does not exist.
  • Critical numbers
    • Definition. Let \(f\) be a function with domain \(D\), and let \(c\) be a number in \(D\). Then \(c\) is a critical number of \(f\), if \(f'(c)=0\), or \(f'(c)\) does not exist.
    • Using this definition, Fermat's theorem reads as follows.
    • If \(f\) has a local extreme value at \(c\), then \(c\) is a critical number of \(f\).

The closed interval method

  • Let \(f\) be a function with domain a closed interval: \(D=[a,b]\).
    • The extreme value theorem is telling us that since \(f\) is defined on a closed interval \([a,b]\), it has a global minimum and a global maximum.
    • Even though Fermat's theorem doesn't find all the local extreme values at once, it decreases their number: for \(f(c)\) to be a local extreme value, we need \(c\) to be a critical number of \(f\).
    • It can also happen that a global extreme value is taken up at one of the endpoints of the closed interval.
  • The closed interval method.
    • To find the global extreme values of \(f\), we can proceed as follows.
    • Step 1. Find the values of \(f\) at its critical numbers.
    • Step 2. Find the values of \(f\) at the endpoints of its domain \([a,b]\).
    • Step 3. The global minimum and maximum values of \(f\) are going to be among the values found in Steps 1 and 2.
  • Exercises. 4.1: 48, 52, 54, 56, 58, 60, 62, 70