(a) Describe fully a sequence of transformations which transforms the graph of y=f(x) to the graph of y=g(x). You should make clear the order in which the transformations are applied.
[5]
The diagram shows the graph of y=g(x).
(b) On the diagram sketch the graph of y=g−1(x) together with any relevant mirror line.
[2]
(c) Find an expression for g−1(x).
[2]
(d) State the range of g−1.
[1]
The function h is defined by
h(x)=x−2for x≥0.
(e) Find the value of g−1h(4).
[1]
(f) Explain why the composite function hg−1 cannot be formed.
Stretch first by scale factor 3 in the y-direction, giving y=3x. Then translate by (−2−5) to obtain y=3x+2−5 on x≥−2.
解法二
思路
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若先平移,竖直位移也会被后续伸缩放大。因此应先下移 5/3,再放大三倍。
答题过程
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Translate first by (−2−5/3), then stretch by scale factor 3 in the y-direction. The result is 3(x+2−5/3)=3x+2−5.
(b)
解法一
思路
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画出镜像直线 y=x,把 g 的端点及曲线上其他点的横纵坐标对调。
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Reflect the increasing branch of y=g(x) across y=x. The endpoint (−2,−5) maps to (−5,−2), and the reflected curve rises to the right. Draw and label the mirror line y=x.
(c)
解法一
思路
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从 y=g(x) 先分离平方根再平方。原函数的输出从 −5 开始。
答题过程
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Writing y=3x+2−5 gives
3y+5=x+2⟹x=(3y+5)2−2.
Thus g−1(x)=(3x+5)2−2 for x≥−5.
(d)
解法一
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反函数的值域等于原函数的定义域。
答题过程
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The range of g−1 is [−2,∞), since the domain of g is x≥−2.
(e)
解法一
思路
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先求 h(4),再将结果代入 (c) 的反函数。
答题过程
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We have h(4)=2, so
g−1(h(4))=g−1(2)=(37)2−2=931.
(f)
解法一
思路
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如果反函数输出了负数,外层函数 h 就无法接收这个值。
答题过程
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The range [−2,∞) of g−1 is not contained in the domain [0,∞) of h. For example, g−1(−5)=−2, where h is undefined. Thus hg−1 cannot be formed on the given domain of g−1.