微積分 計算

Question

f(x)=log| 2x+1 | / 1+2x 。 log| 2x+1 |除1+2x。
以一階與二階檢定法，做表判定極值與反曲點。
詳細解答 謝謝

Answer 1

f' (x) = 2 / (1 + 2 x) 2 - 2 [log| 2 x + 1 |] / (1 + 2 x) 2
= 2{1 - [log| 2 x + 1 |] } / (1 + 2 x) 2 - - - 2 x + 1 = e or - e
x=(e–1)/2=0.86, f(0.86) = 1 - - - 有極值
x= - (e + 1)/2 = - 1.86, f(- 1.86) = - 0.37 - - -有極值
f" (x) = - 8 / (1 + 2 x) 3 - 4 [log| 2 x + 1 |] / (1 + 2 x) 3 + 8[log| 2 x + 1 |] / (1 + 2 x) 3
= 4{[log| 2 x + 1 |] - 2} / (1 + 2 x) 3 - - -
x=(e2 - 1)/2=3.19 , f" (x) = 0 - - 反曲點
x= - (e2 + 1)/2= - 4.19 , f" (x) = 0 - - - 反曲點
x=(e–1)/2=0.86, f" (x) < 0 - - - 相對極大值
x= - (e + 1)/2 = - 1.86, f" (x) > 0 - - - 相對極小值
漸近線: 2x + 1 = 0
lim x- +∞ [log| 2 x + 1 |] / (1 + 2 x) = 0 - - - y = 0 - - - 漸近線
f(10) = (log 19)/ 19 = 0.15
x=(e2 - 1)/2=3.19 , f(3.19) = 0 - - 反曲點
f(2) = (log 5)/5 = 0.32
x=(e–1)/2=0.86, f(0.86) = 0.37 - - - 相對極大值
f(1) = (log 3)/3 = 0.37
f( 0) = (log 1)/ 1 = 0
f(- 0.25) = (log 0.5)/0.5 = - 0.35
lim x- - 0.5+ [log| 2 x + 1 |] / (1 + 2 x) = + ∞
lim x- - 0.5- [log| 2 x + 1 |] / (1 + 2 x) = - ∞ - - - x = - 0.5 - - - 漸近線
f(- 0.75) = (log 0.5)/- 0.5 = - 0.35
f(- 1) = (log 1)/ - 1 = 0
x= - (e + 1)/2 = - 1.86, f (- 1.86) = - 0.37 - - - 相對極小值
f(- 2) = (log 3)/ - 3 = - 0.366
f(- 3) = (log 5)/ - 5 = - 0.32
x= - (e2 + 1)/2= - 4.19 , f(- 4.19) = - 1 - - - 反曲點
f(- 10) = (log 19)/ - 19 = - 0.15
lim x- -∞ [log| 2 x + 1 |] / (1 + 2 x) = 0 - - - y = 0 - - - 漸近線