Page of 5 0 微乙 0-05 班期中考解答和評分標準. (0%) 求 lim 8. 8 ( 8 ) ( ( 8 ) 8 ) ) ( ( ( ) 8 ) 8 ) ) 8 ( ( ( ) 8 ) 8 ) ) 4 8 ) 8 ) Therefore, lim 8 lim 4 8 ) 8 ) 4 4 4 ) ) ) 4/. (0%) 用均值定理說明 tan tan y y 對, y [0, π/4] 均成立 If y, then tan tan y y 0. In general case, we may assume that > y, and, y [0, π/4]. First, we take the differential d d tan sec ( points). By the mean value theorem, we have there eists ξ with y < ξ < such that tan tan y sec ξ( y) (5 points) for all,y [0, π/4]. In the interval [0, π/4], we have sec ξ sec π ( points) 4 so for all,y [0, π/4]. It completes the proof. tan tan y sec ξ y y ( point). (0%) 給定一個方程式 y y 0, 求在點 (, y) (, ) 的切線方程式 Use implicit differentiation. d d (y y ) d d (0) y (y)y y y 0...(8pts) (, y) (, ) y y 0 y.
Page of 5 So the tangent line at (, y) (, ) is y (or y )...(pts) 4. 令 f() 7 7 6 5 4 (a) (0%) 求 f() 在 處之線性逼近 (b) (5%) 以之求 f(0.9) 之近似值 (a) f () 7 6 4 5 0 f () 5, f() f() 5( ) (b) f(0.9 5(0.9 ) 0. 5. (5%) ( 每小題 5%) 計算下列函數之導函數 () sin ( ) () e tan () ln () (sin( )) [%] cos( ) ( ) [%] cos( ) ln. The rest % depends on the detail of your answer. () (e tan ) [%]e tan (tan ) [%]e tan. The rest % depends on the detail of your answer. Another solution: Apply the formula on p45, (f() g() ) [%]g() f() g() f () [%] ln f() f() g() g (). (Here f() e, g() tan ) ( ) ln () Use the quotient rule for derivatives: ln [%] [%]ln The rest % depends on the detail of your answer. 6. (5%) 設計一圓柱形無蓋的杯子 ( 厚度忽略不計 ), 假設在容積固定為 000 立方公分的條件下, 試問使得杯子表面積最小的杯底半徑為何? (%) 需使用極值測試法 (%) let r be the radius of the bottom of the cylinder and let h be the height of the cylinder we have fied volume πr h 000 ( pts) and the surface area A(r) πr πrh ( pts) πr πr 000 πr πr 000 r ( pts)
Page of 5 A (r) πr 000 r πr 000 r ( pts) A (r) 0 iff πr 000 0 iff r 0 ( pts) if r > 0, since πr 000 > 0 and r > 0 we have A (r) πr 000 r > 0 for all r > 0 A(r) is increasing for all r > 0 if 0 < r < 0, since πr 000 < 0 and r > 0 we have A (r) πr 000 r < 0 for all 0 < r < 0 A(r) is decreasing for all 0 < r < 0 we thus conclude that the cylinder has minimum surface area as r 0 ( pts) (if you just say that r is a local minima, you get only pt) ( 另解 ) 本題批改方式為分段給分, 其中 設定區 佔 6 分, 計算區 佔 6 分, 極值測試 佔 分, 每一區詳細配分標註於解法中 從計算錯誤的地方開始之後都不給分, 例如一開始就設定錯誤就不給分 根據考試規則, 只寫出結果者不給分, 常見於 極值測試 中, 直接寫出一階微分或二階微分正負號而沒有計算過程, 或是只畫個簡圖或表格, 都無法判斷是因為題目指定要最小值而把結果直接寫上去, 或是自己算出來的, 這樣的狀況都不給分 設定區 : 設底面半徑為 r, 高為 h, 則 : 體積 V (r) 000 πr h ( pts) 表面積 A(r) πrh πr ( 注意題目為 無蓋 ) ( pts) 將 h 000 πr 帶入 A(r) 得 A(r) πr 000 ( pts) r 計算區 : 求 A(r) 的極值, 即求滿足 A (r) 0 的點 : 極值測試 : 一階測試 :( pts) 故 A(r) 在 r 大於 0 A (r) πr 000 r ( pts) A (r) 0 000 πr 0 r 0 ( pts) A (r) > 0 000 πr > 0 r > 0 A (r) < 0 000 πr > 0 r < 0 遞增, 在 r 小於 0 時遞減, 所以在 r 0 發生最小值 註 : 若只在 r 0 種狀況給 分 這個點上用二階測試, 只能確定他產生 局部極小值, 無法確定他是 最小值, 這 註 : 若在 計算區 中用算幾不等式者, 當作沒有做極值測試, 若都寫對的話給 分 ( ) 7. (5%) 令 y f(). 回答以下各小題 ( 若不存在的話, 須註明不存在 ): (a) y f() 的遞增區間 (, 0] [, ), y f() 的遞減區間 [0, ) (, ] (b) y f() 之極大值 ( 座標 ) (0, ), y f() 之極小值 ( 座標 ) (, ) (c) y f() 的凹向上區間 (, ), y f() 的凹向下區間 (, ) (d) y f() 之反曲點 ( 座標 ) f has no inflection point on R
Page 4 of 5 (e) y f() 所有的漸近線為 y (f) 畫出 y f() 的圖形 For f to be well-defined, we need. Then we can simplify f as Hence we observe that: f() ( ) ( ) f () ( ) f () ( ) (a.) f () 0 for (, 0] [, ) f is increasing on (, 0] [, ) (a.) f () 0 for [0, ) (, ] f is decreasing on [0, ) (, ] Furthermore, f () 0 for 0 or, and from the two observations above, we have (b.) f has a local maimum at (0, ) (b.) f has a local minimum at (, ) And to observe the conveity and concavity. (c.) f () 0 for f is conve on (, ) (c.) f () 0 for f is concave on (, ) Now we observe the inflection point and asymptote. (d) f () 0 for all R f has no inflection point on R (e) f() tends to infinite when tends to is an asymptote of f and f() ( ) 0 as y is an asymptote of f ( ) Finally, we can figure the graph by the above observations. (f) 0 8 6 4 (,) 0 4 (0, ) 6 8 y 0 0 8 6 4 0 4 6 8 0 Grading evaluation:
Page 5 of 5 (a.) If your answer is (, 0), (, ), (, 0], (, ), (, 0), [, ), or (, 0], [, ), you will get points. If your answer only holds one of the two intervals, you will get point. (a.) If your answer is (0, ), (, ), [0, ), (, ), (0, ), (, ], or [0, ), (, ],, you will gent points. Moreover, if your answer contains {}, we also give you points. If your answer only holds one of the two intervals, you will get point. (b.) If your answer is (0, ) or, you will get points. (b.) If your answer is (, ) or, you will get points. (c.) If your answer is (, ) or [, ), you will get points. (c.) If your answer is (, ) or (, ], you will get points. (d) If your answer is or nowhere, you will get points. (e) If your answer is, y, you will get 4 points. If your answer holds one of the two asymptotes, you will get points. (f) If your graph of f is wrong, you have no point. If your graph of f is nearly the true graph, you will get point. And if you figure the two asymptotes, you will get more point; if you mark the local maimum point and local minimum point (you also need to write the coordinates), you will get more point.