Let f (x) have a second-order continuous derivative and f '(x) = 0, LIM (x tends to 0) f' '(x) / |x| = 1, then

Let f (x) have a second-order continuous derivative and f '(x) = 0, LIM (x tends to 0) f' '(x) / |x| = 1, then

f(x) = (1/6)|x^3|
analysis:
If x > 0, f (x) = (1 / 6) x ^ 3, f '(0) = 0, f' '(x) = x, and f' '(x) / |x| = 1, when X - > 0 +
If x < 0, f (x) = - (1 / 6) x ^ 3, f '(0) = 0, f' '(x) = - x, and f' '(x) / |x| = 1, when X - > 0 -
It can be seen that f (x) = (1 / 6) | x ^ 3 | meets all the conditions given in the question

Let f (x) have a second-order continuous derivative, f '(0) = 0, LIM (x tends to 0) f' '(x) / |x| = 1, then Is f (0) the maximum or minimum of F (x)? Why?

LIM (x tends to 0) f '' (x) / |x| = 1
Therefore, near 0) f '' (x) > 0, so the curve is concave, so: F (0) is the minimum of F (x)

Why can the derivative be written as f '(a) = LIM (x → a) f (x) - f (a) / x-a Shouldn't it be written as f '(x) = LIM (△ x → 0) (x + △ x) - f (x) / △ x?

Different symbols mean different

How to calculate LNA LNB

lna-lnb=ln(a/b)
Logarithmic formula

Known function f (x) = ln (1-x) - X / (x + 1) (1) find the minimum value of F (x) (2) if a > 0 b > 0, verify LNA LNB > = 1-B / A

(1) F (x) = ln (1-x) - X / (x + 1) = ln (1-x) - 1 + 1 / (x + 1), domain x < 1 and X ≠ - 1, f '(x) = 1 / (1-x) - 1 / (x + 1) ²= Pass differentiation = x (x + 3) / [(1-x) (1 + x) ²], It can be seen from the definition field that the derivative denominator is always positive. When 0 < x < 1 and x < - 3, the derivative is positive, - 3 < x < - 1 and - 1 < x < 0

It is known that the function f (x) = ln (2-x) + ax is an increasing function in (0,1). (1) find the value range of real number a; (2) If b > 1, verification: ln (B + 2) + lnb-2ln (B + It is known that the function f (x) = ln (2-x) + ax is an increasing function within (0,1). (1) find the value range of real number a; (2) if b > 1, verify: ln (B + 2) + lnb-2ln (B + 1) > - 1 / [b (B + 1)]

1) . f '(x) = 1 / (X-2) + a > = 0, f' (x) is a subtractive function, where f '(1) = - 1 + a > = 0, a > = 1 is obtained
2) Let a = 1, x = 1-T, i.e. 0