課程單元 課程簡介 教學大綱 製作團隊 關鍵詞彙 意見反映
首頁 > > > >

 

\({\rm{Let  }}y = f(x) = x - \frac{{{x^2}}}{6} - \frac{{2\ln x}}{3}\)，\(x > 0\)
Answer the following questions and give your reasons (including computations). Put "None" in the blank if the item asked does not exist.
Find the interval(s) on which \(f\) is increasing. \(\underline {\quad \quad (1\;,\;2)\quad \quad } \)
Find the interval(s) on which f is concave up.  \(\underline {\quad \quad (0\;,\;\sqrt 2 )\quad \quad } \) 【台大數學系102年度期中考】

詳解：\(f'(x) = 1 - \frac{x}{3} - \frac{2}{{3x}}\)，\(x > 0\)

\(f'(x) = 1 - \frac{x}{3} - \frac{2}{{3x}}\; = \frac{{3x - {x^2} - 2}}{{3x}} = \frac{{( - x + 1)(x - 2)}}{{3x}}\quad \)，\(x > 0\)

\(x\)   0   1   2  
\(f'(x)\) x \(\vdots \) - \(\vdots \) + \(\vdots \) -
\(f(x)\) x \(\vdots \) ↘ \(\vdots \) ↗ \(\vdots \) ↘

 

臨界值\(x = 0\;,x = 1\;,x = 2\)

\(f''(x) =  - \frac{1}{3} + \frac{2}{{3{x^2}}}\; = \frac{{ - {x^2} + 2}}{{3{x^2}}} = \frac{{(\sqrt 2  + x)(\sqrt 2  - x)}}{{3{x^2}}}\)，\(x > 0\)

二階臨界值\(x = 0\;,x = \sqrt 2 \;,x =  - \sqrt 2 \)

\(x\)   0   \(\sqrt2\)  
\(f'(x)\) x \(\vdots \) + \(\vdots \) -
\(f(x)\) x \(\vdots \) \(\cup\) \(\vdots \) \(\cap\)


創用 CC 授權條款
微積分一 calculus I 由CUSTCourses 李柏堅製作，以創用CC 姓名標示-非商業性-禁止改作 3.0 台灣 授權條款釋出