题目
The curve has equation .
(a) Find the equations of the asymptotes of .
(b) Find the coordinates of the stationary points on .
(c) Sketch .
(d) Sketch the curve with equation and find in exact form the set of values of for which .
题目中文翻译
曲线 的方程为 。
(a) 求 的渐近线方程。
(b) 求 上驻点的坐标。
(c) 画出 的草图。
(d) 画出方程为 的曲线的草图,并用精确形式求出满足 的 的取值集合。
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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多项式除法得到斜渐近线,分母零点不能约消。
答题过程
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Hence the asymptotes are and .
(b)
解法一
思路
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用商法则并因式分解分子,两个驻点都在定义域内。
答题过程
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Setting the numerator to zero gives or . Substitution yields
The derivative changes from positive to negative at and from negative to positive at , giving a maximum and minimum respectively.
(c)
解法一
思路
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左支非正并在原点平滑接触横轴,右支有正的最低点。余项符号决定两支与斜渐近线的相对位置。
答题过程
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Label and . For , sketch a branch below the oblique asymptote, rising from to touch the -axis at its maximum and then falling to as .
For , sketch a branch above the oblique asymptote, falling from to the minimum and then rising to .
Both outer tails approach the oblique asymptote, and is the only axis intersection.
(d)
解法一
思路
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反射整条左支,原点仍为光滑最低点。解绝对值不等式时分别在竖直渐近线两侧处理分母符号。
答题过程
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For the absolute-value graph, reflect the branch with in the -axis, retaining the branch with . The origin becomes a smooth minimum, the right minimum remains , and both sides tend to at .
The outer asymptotes are on the left and on the right.
For , the original function is nonpositive, so the inequality is equivalent to . Multiplication by the negative denominator reverses the sign:
The roots are , both less than . Hence this gives
For , the original function is positive, and multiplication by the positive denominator gives
Its roots are , both greater than . Combining the two cases,