![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
F (x) = x ^ 2 - ax + ln (x + 1), a belongs to R 0 F extremum when a = 2 If the function f (x) always has f prime (x) greater than x in the interval (0,1), find the value range of real number a The word "0" doesn't mean The speed of copying is fast
F '(x) = 2x - 2 + 1 / (x + 1) = 0x = √ 2 / 2 or - √ 2 / 2y1 = 1 / 2 - √ 2 + ln (√ 2 / 2 + 1) y2 = 1 / 2 + √ 2 + ln (- √ 2 / 2 + 1) the extreme point is (√ 2 / 2,1 / 2 - √ 2 + ln (√ 2 / 2 + 1)) (- √ 2 / 2,1 / 2 + √ 2 + ln (- √ 2 / 2 + 1)) f' (x) = 2x - A + 1 / (x + 1) > x, a ≤ 1
RELATED INFORMATIONS
- 1. f(x)=ln(x+1),lim(x->0)
- 2. LIM (x tends to positive infinity) cosx / [e ^ x + e ^ (- x)] =?
- 3. Prove LIM (X -- > x0) 2 ^ x = 2 ^ x0 by definition
- 4. lim x→∞ x^2-6x+8/x^2-5x+4 Wrong. It's not infinity. It's four
- 5. Help find a limit: LIM (x ^ 2-6x + 8) / (x ^ 2-5x + 4) x → 4
- 6. 1.5 find the limit LIM (x → 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4) 1.5 find the limit LIM (x → 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4)
- 7. Finding limit LIM (x ^ 2-6x + 8) / x ^ 2 + X-6
- 8. Limit: LIM (x - > 4) (x ^ 2-6x + 8) / (x ^ 2-5x + 4)
- 9. Detailed process of LIM (x → 1) lncos (x-1) / (1-sin (π X / 2))
- 10. Find Lim sin (x-1) / (x ^ 2-1) where X - > 1
- 11. Given function f (x) = ln (AX) / (x + 1) - ln (AX) + ln (x + 1) Given the function f (x) = ln (AX) / (x + 1) - ln (AX) + ln (x + 1), (a is not equal to 0 and R) 1. Find the definition field of function f (x) 2. Find the monotone interval of function f (x) 3. When a > 0, if there is x such that f (x) ≥ ln (2a) holds, the value range of a is obtained
- 12. Ln x = ax has several real roots (a > 0)
- 13. Find Lim x → 0 (Tan 2x SiNx) / X and lim x → 0 (COS x-cos 3x) / x ^ 2
- 14. Is the square of LIM x → 0 (cosx-cos3x) / X equal to? Lim x → 0 (cosx-cos3x) / x square, please write the steps!
- 15. The limit of tanx-6 / secx + 5 when x tends to Π / 2
- 16. lim(x→0)[(secx)^2-1] /(1-cosx)
- 17. Given function, find Lim f (x) Given the function y = {x + 1, x < 1, Lim f (x) =? 1/x,x≥1 x→0 A. 1 b.0 C. No D.2
- 18. Given the function f (x) = 2ln3x + 8x, the value of LiMn →∞ f (1 − 2 △ x) − f (1) △ x is () A. 10B. -10C. -20D. 20
- 19. Let f (x) be defined in a neighborhood of x = a, then a sufficient condition for f (x) to be differentiable at x = a is? A.lim (x approaches 0) [f (a + 2H) - f (a) Dlim (x tends to be 0) H [f (a + 1 / h) - f (a)] exists. Explain in detail why abd is not right (especially d)? A. LIM (H tends to 0) [f (a + 2H) - f (a + H)] / h exists, while LIM (H tends to 0) [f (a + H) - f (A-H)] / 2H exists C. LIM (H tends to 0) [f (a) - f (A-H)] / h exists. ABC option was forgotten just now, now add it. D is changed to dlim (H approaches infinity) H [f (a + 1 / h) - f (a)]
- 20. LIM (x → 0 +) lncotx / LNX is used to find the limit