School

Western UniversityDepartment

MathematicsCourse Code

MATH 0110A/BProfessor

Chris BrandlStudy Guide

QuizThis

**preview**shows pages 1-2. to view the full**7 pages of the document.**MATH 127 Fall 2012

Assignment 2 Solutions

1. For each part, sketch all of the given functions on the same set of axes.

(a) (i) f(x) = √x(ii) g(x) = −√−6x(iii) h(x) = −2√2−x−1

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y=f(x)

y=g(x)

y=h(x)

(b) (i) f(x) = e−x(ii) g(x) = −e−x+1 (iii) h(x) = 1

e−x+e

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y=f(x)

y=g(x)

y=h(x)

(c) (i) f(x) = 1

x(ii) g(x) = 1

x+ 1 + 1 (iii) h(x) = 4

x

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y=g(x)

y=h(x)

y=f(x)

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2. Sketch each function.

(a) f(x) = |x+ 1|+|x−1|(hint: this can be expressed as a piecewise function, show

the work in getting this piecewise function)

We have that x+ 1 ≥0 when x≥ −1 and that x−1≥0 when x≥1. We now

consider the three intervals x < −1, −1≤x < 1, and x≥1.

For the ﬁrst interval x < −1 we have that x+ 1 <0 and x−1<0 so we get that

f(x) = |x+ 1|+|x−1|=−(x+ 1) −(x−1) = −x−1−x+ 1 = −2x.

In the second interval −1≤x < 1 we have that x+ 1 ≥0 and x−1<0 so we

get that f(x) = |x+ 1|+|x−1|= (x+ 1) −(x−1) = x+ 1 −x+ 1 = 2.

In the third interval x≥1 we have that x+ 1 ≥0 and x−1≥0 so we get that

f(x) = |x+ 1|+|x−1|= (x+ 1) + (x−1) = x+1+x−1 = 2x.

Thus we have that f(x) = |x+ 1|+|x−1|=

−2xif x < −1

2 if −1≤x < 1

2xif x≥1

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(b) f(x) = 1

x2−4x+ 6 (hint: ﬁrst look at the graph of x2−4x+ 6)

First consider the graph of y=x2−4x+ 6. Completing the square we can re-

write this as y= (x−2)2+ 2 so we see that this is a parabola shifted 2 to the

right and 2 up. The graph of y= (x−2)2+ 1 goes through the point (2,2) and

approaches +∞for large negative and positive values of x. From this we see that

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