Suppose that f(x) is positive everywhere. Let g=1 / f. If f has a relative maximum at x=a, what

Suppose that f(x) is positive everywhere. Let g=1 / f. If f...

Key Concepts

Effect of Inversion on Extrema

When applying the reciprocal transformation to a positive function, the nature of local extrema is reversed. This means that if the original function f has a relative maximum at a particular point, then the reciprocal function g = 1/f will have a relative minimum at that same point. This inverse relationship is due to the fact that the reciprocal function decreases as the original function increases and vice versa.

Reciprocal Transformation

This concept refers to the process of forming a new function as the reciprocal of an existing function—that is, creating g(x) = 1/f(x). When the original function is positive everywhere, as given, the transformation is well-defined and continuous. Understanding how this transformation affects the properties of the original function is crucial, particularly when examining behavior near extrema.

Relative Extrema

Relative (or local) extrema are points in the domain of a function where the function values are locally either the highest or the lowest relative to nearby points. A relative maximum is a point where the function value is greater than or equal to the values of the function in nearby points, while a relative minimum is where it is less than or equal to the values in its vicinity.

Transcript

00:01 Let's assume that a function of two variables, f of x, y, has a relative maximum at x, 0, y.

00:10 We will show that the function of one variable defined as g of x equal f of x y0, that is fixing the second variable at the value y zero in the function f.

00:23 That function of one variable has a relative maximum at x equal x zero.

00:28 And at the same time the function of one variable defined as h of y equals f of x0 y that is fixing now the first variable to the value x0 for the function f that function of one variable has a relative maximum at y equals y0 so first we're going to prove that g of x define as f of x y0 has a relative maximum at x equal x0.

01:17 For that we use the hypothesis that the function f of two variables as a relative maximum at x0 y0.

01:25 So f has a relative maximum at x0 y0 means by definition that derksis a disk let's call it d center at x0y 0 such that for every point in the disk let's say xy on d is true that or it happens that f of x0 is greater than or equal to f xy xy that is the image of the function f at the point x0 .0 is greater than or equal to every image of every point into disk d.

03:07 So there is at least one disk that has this property.

03:11 The disk is centered at x0.

03:18 So let's say that the disk centered x0 or zero is defined as the points in the plane such that.

03:36 That the distance between x y and x zero is less than a radius less than or equal.

03:48 So this is the square root of x minus x0 square plus y minus y0 square.

03:58 That's a distance between x y and x0 with zero that is less than or equal to r.

04:05 For some radius are positive.

04:19 Well, now we are going to prove that g of x is, has a relative maximum at x0.

04:28 So we are gonna take any point in the interval of radius r and center at x0.

04:39 So let x be any point, sorry here, let xb any point in the interval x0 minus r x0 plus r that is the close interval of length to r center at x0 so if the point x or the value x is in this interval then we know that the distance between x and x0 that is the absolute value of x minus x0 is less than or equal to r the equality is included because the interval is closed so we are taking any point in this close interval center at x0 of radius radius r and and we are going to see what happened with the function g at the point and this implies then that the square of this expression x minus minus x0 absolute value, that square of that is less than or equal to r square.

06:15 Because the square, the quadratic function, is increasing for the positive numbers.

06:21 And these values here are positive or zero.

06:27 And the real numbers, we know that the absolute value square is the same as the value square.

06:32 So we can write this without absolute value is the same thing.

06:36 And now this implies that the absolute x minus x0 square plus y0 minus y0 square because this is zero we are not adding anything at all.

06:56 So this is for inequality is the same as the previous one...

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