3.3.13.3.23.3.3例4设o<a<b证明bb-ab-a<lnIbaa证明由于o<a<b,故在[a,b]上，函数ln显然满足中值定理的条件，所以存在一点E（a,b）使得)==(b-alnb - ln a = (ln a)la=(b - a)但11司bwa即有所证的结果取a=>0,b=+1可得1In1a+1r返回全屏关闭退出11/34
3.3.1 3.3.2 3.3.3 ~ 4 0 < a < b, y² b − a b < ln b a < b − a a . y² du 0 < a < b, 3 [a, b] þ, ¼ê ln x w,÷v¥½n^ , ¤±3: ξ ∈ (a, b) ¦ ln b − ln a = (ln x) 0 |x=ξ(b − a) = 1 ξ (b − a) 1 b < 1 ξ < 1 a =k¤y(J. a = x > 0, b = x + 1 1 x + 1 < ln 1 + 1 x < 1 x . 11/34 kJ Ik J I £ ¶ '4 òÑ
3.3.33.3.13.3.2例5证明恒等式元[α] ≤1arcsin + arccos 2证明 命 f(α)= arcsin + arccos a.则 f(α)三 o.所以arcsin + arccos 三 c(常数).将α=0 代入,即得 c=例 6 设 f(α)在[a,b] 连续, 在 (a,b) 可导,且 f(a) = f(b) = 0. 求证: 存在E (a,b) 使得f'(s) - εf() = 0.证明 令 g(c) = e-rf(c). 则 g(c)在[a,b] 连续,在 (a,b) 可导,且g(a) = g(b) = 0. 根据 Rolle 定理, 存在 E (a, b) 使得 g(s) = 0, 即-te-" f(E) +e-" f'(E) = 0,也就是 f(E) - f(S) = 0.返回全屏关闭退出II12/34
3.3.1 3.3.2 3.3.3 ~ 5 y²ð ª arcsin x + arccos x = π 2 , |x| 6 1 y² · f(x) = arcsin x + arccos x. K f 0 (x) ≡ 0. ¤± arcsin x + arccos x ≡ c (~ê). ò x = 0 \, = c = π 2 . ~ 6 f(x) 3 [a, b] ëY, 3 (a, b) , f(a) = f(b) = 0. ¦y: 3 ξ ∈ (a, b) ¦ f 0 (ξ) − ξf(ξ) = 0. y² - g(x) = e −1 2 x 2 f(x). K g(x) 3 [a, b] ëY, 3 (a, b) , g(a) = g(b) = 0. â Rolle ½n, 3 ξ ∈ (a, b) ¦ g 0 (ξ) = 0, = −ξe−1 2 ξ 2 f(ξ) + e −1 2 ξ 2 f 0 (ξ) = 0, Ò´ f 0 (ξ) − ξf(ξ) = 0. 12/34 kJ Ik J I £ ¶ '4 òÑ