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Mathematics

MATA30H3

Sophie Chrysostomou

Winter

Description

University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT A30Y
Solution to Assignment #4
A. Homework problems from the lectures:
√
1. (a) Let f(x) = sinx, g(x) = 1 − x . Find f ◦ g(x) and g ◦ f(x) and their domains.
√ 2
solution : (f ◦ g)(x) = f(g(x)) = sin(g(x)) = sin( 1 − x ) with domain the set
{x ∈ R|x ∈ domg and g(x) ∈ domf}.
The domain of g is:
domg = {x ∈ R|1 − x ≥ 0}
1 − x ≥ 0 ⇐⇒ x ≤ 1 ⇐⇒ |x| ≤ 1 ⇐⇒ −1 ≤ x ≤ 1 ∴ domg = [−1, 1]
Since the domain of f(x) = sinx is R, then g(x) ∈ domf for all x ∈ domg. Thus,
domf ◦ g = [−1, 1].
p p 2 √
(g ◦ f)(x) = g(f(x)) = 1 − f(x) = 1 − sin x = cos x = |cosx| with domain the
set
{x ∈ R|x ∈ domf and f(x) ∈ domg}.
√
Since f(x) = sinx then domf = R. Since g(x) = 1 − x = [−1, 1] and −1 ≤ sinx ≤ 1
then f(x) ∈ domg for all x ∈ domf = R. Thus, domg ◦ f = R.
x − 2x + 3
(b) If h(x) = 2 2 ﬁnd f and g such that f(x) ▯= x and g(x) ▯= x and h = f ◦g.
(x − 2x + 4)
solution :
x − 2x + 3 (x − 2x + 1) + 2 (x − 1) + 2
h(x) = 2 2 = 2 2= 2 2.
(x − 2x + 4) ((x − 2x + 1) + 3) ((x − 1) + 3)
2 x + 2
Let g(x) = (x − 1) and f(x) = 2, then f ◦ g = h.
(x + 3)
1 2. A) From the graph of cosx (in blue), we get the graph of secx (in green):
5
4
3
2
1
▯π 0 π
▯1
▯2
▯3
▯4
We need a one to one function so deﬁne f(x) = secx with domain [0,π/2) ∪ [π,3π/2)
and range (−∞,−1] ∪ [1,∞). Its graph is in bold green color. Then f is one to one
and
y = f −1 (x) = sec −1 x = arcsecx for x ∈ (−∞,−1]∪[1,∞) and range [0,π/2)∪[π,3π/2).
To get the graph of f −1 we get the mirror image of the graph of f at the mirror y = x:
The graph of arcsec x
π
▯5 ▯4 ▯3 ▯2 ▯1 0 1 2 3 4
B) Similarly From the graph of tanx (in blue), we get the graph of cotx (in green):
2 5
4
3
2
1
▯π 0 π
▯1
▯2
▯3
▯4
we deﬁne f(x) = cotx with domain (0,π) and range R. Then:
y = f −1(x) = cot1 x = arccotx for (x ∈ R) and range (0,π).
−1
To get the graph of f we get the mirror image of the graph of f at the mirror y = x:
The graph of arccot x
π
▯5 ▯4 ▯3 ▯2 ▯1 0 1 2 3 4
" ! #
√
3
3. Simplify sin arccos 2 + arctan(1) .
solu"ion : √ ! #
3 π π
sin arccos + arctan(1) = sin +
2 6 4
π π π π .
= sin 6 cos 4 + cos6 sin 4
√ √
1 1 3 1 1 + 3
= ▯√ + ▯√ = √
2 2 2 2 2 2
3 Stewart. Section 1.6:
4 Stewart. Section 1.6:
5 Stewart. Section 1.6 & Appendix D:
6 C. Questions from topics in Calculus Review Manual (CRM):
1. Read pages 22-23.
2. Page 24: # 4, 14.
(x − 5x + 4)(x + 2)
(4.) Find all x satisfying ≥ 0
(x + 3)(2x + 1)
2
solution : Let F(x) = − 5x + 4)(x + 2)
(x + 3)(2x + 1)
A. & B. ) Factor and simplify:
(x − 5x + 4)(x + 2(x − 4)(x − 1)(x + 2)
F(x) = 2

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