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It is known that the function y = f (x) is an odd function defined on R, and when x
(1) If x > 0, then there is f (x) = - f (- x) = - 1-2 ^ (- x) when there is - x0, so its analytical formula on R is: F (x) = - 1-2 ^ (- x), (x > 0) = 0, (x = 0) = 1 + 2 ^ x (2) monotonic increasing interval is (- infinite, 0) and (0, + infinite) (3) f (1 + x) + F (2x) > 0f (1 + x) > F (- 2x), so there is 1 + x > - 2x > 0 or 0 > 1 + x > - 2x or 1 + x0, that is
F (x) = x ^ 3-x find the derivative of x = 2 Want process
f(x)=x^3-x
Then f '(x) = 3x ²- one
So f '(2) = 3 * 2 ²- 1=12-1=11
When h tends to 0, (f (x0 + 2H) - f (x0 + H)) H is equal to the derivative of F (x + H) High number! Seeking education
(f(x0+2h)-f (x0+h)) /h
Take the derivative of H with lobida's law
= (2f'(x0) -f'(x0))/1
=f'(x0)
If the derivative of F (x0) is 3, Lim delta x tends to 0: H is equal to 12 on the ratio of F (x0 + H) - f (x0-3h) It's unscientific to calculate the ratio of F (x0 + H) - f (x0-3h) to h by the condition of F (x0). How to calculate it? Shouldn't f (x0) be the condition of F (x0 + H) - f (x0) to h?
same
[f(x0+h)-f(x0-3h)]/h=4{[f(x0-3h) +4h]-f(x0-3h)]/4h
F (x0-3h) is equivalent to f (x) in the formula, and 4H is equivalent to △ X in the formula
When h approaches 0, f (x0-3h) = f (x0)
It is proved that if two derivable functions f (x) and G (x) satisfy that f (0) = 0, G (x) = 0 and their derivatives exist, G (x) is not 0, then f (x) / g (x) The limit of is f (x) derivative / g (x) derivative
This is the law of lobita. But you have omitted one condition: Lim f '(x) / g' (x) exists (or is infinite)
There is a process in the book. It comes out immediately after using Cauchy mean value theorem
I hope it can help you. If you solve the problem, please click the "select as satisfactory answer" button below,
How did RN (x) of Lagrange remainder of Taylor formula come from? I mean, how to find RN (x), that is, how to use its expansion
If the function f (x) has derivatives up to order n + 1 in the open interval (a, b), when the function is in this interval, it can be expanded into a sum of (X-X.) polynomials and a remainder: F (x) = f (X.) + F '(X.) + F' '(X.) + F' '(X.) / 2! • (X-X.) ^ 2, + F' '' (X.) / 3! • (X-X.) ^ 3 +... + F (n) (x
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