Porisma

Worked solution

Invert a piecewise function with a gap in its range

The function gg is defined by

g:x↦{x−1,x≤2,x+1,x>2.g:x\mapsto\begin{cases}x-1,&x\leq2,\\\sqrt{x+1},&x>2.\end{cases}

(i) Sketch the graph of y=g(x)y=g(x). marks

(ii) Explain why the inverse function, g−1g^{-1}, exists. Find g−1g^{-1} in the similar form. marks

Another function hh is defined by

h:x↦ex+1h:x\mapsto e^{x+1}, x∈Rx\in\mathbb R.

(iii) Find hg(x)hg(x) for x<0x<0. mark

(iv) Find the range of the composite function hghg for x<0x<0. mark

[2010/RVHS/Promo/8]

CJC · Review Problems · Question 9

Answer

g−1(x)=x+1g^{-1}(x)=x+1 for x≤1x\leq1, and x2−1x^2-1 for x>3x>\sqrt3. For x<0x<0, hg(x)=exhg(x)=e^x with range (0,1)(0,1).

(i)

Draw the line y=x−1y=x-1 for x≤2x\leq2 with included endpoint (2,1)(2,1), then the square-root curve y=x+1y=\sqrt{x+1} for x>2x>2 with excluded endpoint (2,3)(2,\sqrt3). Thus Rg=(−∞,1]∪(3,∞)R_g=(-\infty,1]\cup(\sqrt3,\infty).

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(ii)

Every horizontal line meets the graph at most once: both branches increase, and their output intervals are disjoint. Hence gg is one-to-one and its inverse exists. Invert the branches separately to obtain g−1:x↦{x+1,x≤1,x2−1,x>3.g^{-1}:x\mapsto\begin{cases}x+1,&x\leq1,\\x^2-1,&x>\sqrt3.\end{cases} The gap (1,3](1,\sqrt3] is outside the inverse domain.

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(iii)

For x<0x<0, use the first branch g(x)=x−1g(x)=x-1. Therefore hg(x)=h(x−1)=e(x−1)+1=exhg(x)=h(x-1)=e^{(x-1)+1}=e^x.

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(iv)

As xx ranges over (−∞,0)(-\infty,0), exe^x ranges over (0,1)(0,1). Neither endpoint is attained, so Rhg=(0,1)R_{hg}=(0,1).

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Compare your sketch

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Piecewise graph: line y=x−1 for x≤2 ends at the filled point (2,1); square-root curve y=√(x+1) for x>2 begins at the open point (2,√3). Both branches increase, with disjoint output intervals (−infinity,1] and (√3, infinity).
Piecewise graph: line y=x−1 for x≤2 ends at the filled point (2,1); square-root curve y=√(x+1) for x>2 begins at the open point (2,√3). Both branches increase, with disjoint output intervals (−infinity,1] and (√3, infinity).
Piecewise graph: line y=x−1 for x≤2 ends at the filled point (2,1); square-root curve y=√(x+1) for x>2 begins at the open point (2,√3). Both branches increase, with disjoint output intervals (−infinity,1] and (√3, infinity).

Common mistakes

The second inverse branch begins strictly above √3, leaving a genuine gap in the inverse domain.

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