Proof of functional concave and convex by using quadratic Derivative

Source: Internet
Author: User

Many people actually know that the secondary derivative of the function can be used to judge the concave and convex of the function, but many people forget how to prove it. Here I will prove it again.

Proof: If f (x) is continuous in (A, B) and can be imported twice, if f'' (x)> 0, the function is concave. If f (x) is greater than 0, the function is convex.


First, we will give several theorems and explanations.

Description of Function Concave and convex:

Function f (x) is continuous in (a, B). For any A <X1 <X2 <B, X0 = (X1 + x2)/2,

If F (x0) <(f (X1) + f (X2)/2, the function (up) is considered concave; if F (x0)> (f (X1) + f (X2)/2, the function (up) is considered convex.

A Theorem to be used in the process of proof is given: The mean value theorem of Laplace.

If function f (x) is continuous in [a, B] In (a, B), at least one e exists, and a <e <B causes f' (E) = (F (B)-f (a)/(B-)

The following describes our initial problems:

Any f (x) on two points x1, x2 makes a <X1 <X2 <B, X0 = (X2 + X1)/2, then the x0-x1 = x2-x0, let the x0-x1 = x2-x0 = H, then apply two Laplace Mean Value Theorem in (x1, X1 + H) and (x2-h, X2) respectively

F' (E1) = (f (x0)-f (X1)/H (1)

F' (E2) = (f (X2)-f (x0)/H (2)

E1 is in the range of (x1, X0) and E2 is in the range of (x0, X2 ).

Let (1) and (2) Multiply H to the left, and then subtract to (f' (E2)-f' (E1) H = f (X2) + f (X1)-2 (f (x0) (3)

Then, we use the mean value theorem of a Laplace in (E1, E2) to obtain the following result:

F' (E2)-f' (E1) = f'' (E) (e2-e1) Where E is within the range of (E1, E2, then, if the value of F ''(e) is greater than 0, the left side of (3) is greater than 0. Likewise, the right side of (3) can be obtained.

F (x0) = f (X1 + x2)/2) <(f (X2) + f (X1)/2, combined with the convex theorem of the function, we can obtain that the original function is concave, And the convex proof is similar .. No proof.

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