Subject: Calculus

Extreme Value Theorem

A theorem that categorically state that every continuous function on the interval [a,b] attains its absolute maximum and minimum values on the same interval [a,b] is called as the Extreme Value Theorem. Formally, extreme value theorem has the following definition: If a function f is continuous on the closed interval [a,b], then f has an absolute maximum value and an absolute minimum value on [a,b].
So, this leads us to a step-by-step method in finding the extreme values of a function on a specific interval. Of course extreme values are known as absolute maximum and absolute minimum which can be determined by following the steps below.

  • Solve for the critical values of the function f.
  • Find the function values at the critical numbers of f on the interval (a,b)
  • Find the values of f(a) and f(b).
  • The largest of the values obtained in the second step and third step is the absolute maximum value of f. On the other hand, the smallest of the values obtained is the absolute minimum value of f.

So, using the steps above which mainly revolves around the principles of the extreme-value theorem, we can find the absolute extrema of any function along any given interval. To see this, let's try say, a function f = x^3 + x^2 - x + 1 and find the absolute extrema of this function along the interval [-2, 0.5] by using the extreme value theorem.

Now, according to the steps outlined above, let us first find critical points of this function on the interval. Thus,

FIRST STEP: We differentiate f such that

f' = 3x^2 + 2x -1

Setting this to zero and solve for x,

f' = 3x^2 + 2x -1 = 0

We find the critical values of f to be \frac{1}{3} and -1. Obviously, \frac{1}{3} and -1 is located between the points [-2, 0.5]. Thus, these critical values are critical points of the function along the interval.

SECOND STEP: We have successfully solved for the critical values of the function f along the interval [-2, 0.5] and it happens so that all the critical values coincidentally located in the said interval. Now, the second step tells us to find the function values at the critical numbers of f on the interval (a,b). This can be done by just substituting the critical values x =\frac{1}{3}, -1, to the function. Thus,

f(\frac{1}{3}) = (\frac{1}{3})^3 + (\frac{1}{3})^2 - (\frac{1}{3}) + 1

f(\frac{1}{3}) = \frac{22}{27}


f(-1) = (-1)^3 + (-1)^2 - (-1) + 1

f(-1) = 2.

THIRD STEP: Well, this step is quite easy. Let us just find the values of f at the endpoints of the interval [a, b]. Meaning, we will find f(a) and f(b). Then,

f(a) = f(-2) = (-2)^3 + (-2)^2 - (-2) + 1

f(a) = -1

On the other hand,

f(b) = f(0.5) = (0.5)^3 + (0.5)^2 - (0.5) + 1

f(b) = \frac{7}{8}

FOURTH STEP: This step calls nothing more than just a mere comparison of the results we have obtained in previous steps. Now, the second and third steps produced values which are \frac{22}{27}, 2, \frac{7}{8} and -1 . Among these values, the greatest is 2 obviously and the lowest is -1.

Thus, we have now the values for the absolute extrema of the function f along the interval [-2, 0.5]. Correspondingly, the absolute maximum value is 2 while the absolute minimum value is -1. Alternatively, we can check and verify our results graphically. By looking at the figure at the right, we can see the plot of the function f and by just one look, we can really tell that the graphical result agrees with our calculated result.

Thus, we have just completed another topic in Calculus which is the extreme-value theorem. If we have to express the theorem in a few words, it would be something like; the maximum and minimum values of a function in a given interval, is located along that interval also. Sounds new? Yeah, but its true...

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