MATH 148 Fall 2006 Quiz 4, alternative solutions
1. Let f.x/D1
p2๎x2.
To show that a number ybelongs to the range of fmeans that you have to find a numberxin
the domain of ffor which yDf.x/. That means that you have to solve the equation yDf.x/
for x.
When yD1this means solving the equation
1
p2๎x2D1
for x. Clearing fractions, we get p2๎x2D1. Squaring both sides, we get 2๎x2D1or x2D1.
This means that xD1or xD๎1, i.e., f.1/Df.๎1/D1. Therefore 1belongs to the range of
f, since both 1and ๎1belong to the domain of f.
To show that a number ydoes not belong the the range of fmeans showing that there is no
number xin the domain of ffor which yDf.x/.
So when we have to show that 1=2is not in the range of f, this amounts to showing that the
equation
1
p2๎x2D1
2
has no solutions.
One way to find the range of a function f.x/is to solve the equation f.x/Dyfor xand
identify the domain of the resulting expression.
Here 1
p2๎x2Dy
implies that 1Dyp2๎x2or 1Dy2.2๎x2/. Therefore,
2๎x2D1
y2H) x2D2๎1
y2H) xDห
s2๎1
y2:
For xto be a real number, the expression under the square root sign must be zero or positive, i.e.,
2๎1=y2๎0H) 2๎1=y2H) y2๎1=2. Since yDf.x/D1=p2๎x2>0we
conclude that y๎1=p2Dp2=2. Therefore the range of fis the unbounded closed interval
ลp2=2;1/.
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