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Local Minimum and Local Maximum
Motivation: From the derivative of a function, it can be known that the function is monotonically increasing or decreasing. And from 2nd order derivative, it can reveal the information that the function is convex or concave. Those properties can be found by checking if the derivative or 2nd order derivative is larger or less than 0. How about the situation when the derivative is 0 at given point ? Let’s consider a simple diagram as below:
The function is monotonically increasing while progressing from the left to the right before hitting the green line. After passing the green, the function starts decreasing. When the function is monotonically increasing, its derivative is larger than 0; when the function is monotonically decreasing, its derivative is less than 0. Thus, can we say the derivative will be equal to 0 when it reaches the maximum point ? That sounds correct. But the converse statement is true ? In other words, if the derivative at a given point is 0, does the function have maximum value over there? Check the following two diagrams:
The left one is with minimum value when the slope of the tangent is 0; the right one is neither maximum nor minimum when the slope of the tangent is 0.
How about checking if it is convex or concave while checking if the function is increasing or decreasing to find the maximum or minimum? Think about this: minimum happens when the slope of the tangent changes from negative to positive; that means the 2nd order derivative is larger than 0 so that the 1st order of derivative is monotonically increasing. Similarly, maximum happens when the slope of the tangent changes from positive to negative. So, this can be a way to find maximum and minimum.
Definition ( Critical Point ) Let f(x) be a differentiable function. If f’(c) = 0, then x=c is known as a critical point of f(x).
Theorem ( Local Maximum and Local Minimum ) Let f(x) be a function such that its 1st order derivative and 2nd order derivative exist. And it has critical point when x=c, i.e., f’(c)=0 . Then (1) When f”(c) > 0 , f(x) has local minimum at x=c . (2) When f”(c) < 0 , f(x) has local maximum at x=c . (3) When f”(c)=0, the information is not enough to determine it is local minimum, or local maximum, or else. It needs to be determined by higher order derivative.
Proof: Proof of (1) and (2) are just summarizing what we know from previous sections: f’(c) determines if f(x) is monotonically increasing or decreasing; f”(c) determines if f(x) is convex or concave. For more formal proof, you can just use
f is convex f(d) f(c) + f’(c)(d-c) for any c, d in the domain of f
f is concave f(d) f(c) + f’(c)(d-c) for any c, d in the domain of f
It has been proved in the previous section. The problem is: when the 2nd order of derivative is 0, we are not able to tell whether it is convex or concave. Actually, the result here is just a trivial result with Taylor series expansion to approximate a function:
f(x) = f(c)+ f’(c)(x-c) + f”(c)(x-c)2 + … + f(n)(c)(x-c)n + f(n+1)(b)(x-c)n+1 and if you use those terms up to f(n)(c)(x-c)n to approximate f(x), the error is f(n+1)(b)(x-c)n+1 , where b is between x and c.
So, if you find the first (n-1)th order derivatives are 0, and f(n)(c) is the first one that is non-zero, then
when n is even, f(c) is local maximum or minimum; when n is odd, f(x) at x=c is just an inflection point ( saddle point ).
The reason is : (x-c)n is always positive when n is even no matter when x is larger or less than c; but when n is odd, that is not the case.
Taylor series expansion will be introduced later. Here, we only use the result to state how to find local maximum and minimum.
The reason why we call it “local minimum” or “local maximum” is that this “maximum” or “minimum” is only local to its neighborhood points. For example, in the following diagram, point A and C are local maxima. But there are some other points that are with larger values than A and C.
Example: f(x)=x3 -2x2 -4x+1
f’(x) = 3x2 – 4x -4 = (3x+2)(x-2) f”(x) = 6x – 4
f’( )=0, f’(2)=0 and f”( )=-8 < 0 , f”(2)=8 > 0 So, f(x) has local minimum at x=2, and local maximum at x= .
Example: f(x)=x3 f’(x)=3x2 f”(x)=6x f(3)(x)=6
f’(0)=0 . But if we draw the diagram directly, we have
At x=0, it is neither local maximum or minimum; it is known as “inflection point”.
Example: f(x)=x4-1
f’(x)=4x3 f”(x)=12x2 f(3)(x) = 24x f(4)(x) = 24
f’(0)=0, f”(0)=0 . But the diagram of f(x) is as follows:
At x=0, f(x) has minimum.
The previous examples just show that it is not sufficient to determine if it is local maximum or minimum when f’(x)=0 and f”(x)=0 . Keep this result in mind so that you will not make mistake. |
