Download Honors Calculus I

HW 2
2–4
Let f (x) = (x − 1)2 − 1.
(a) Explain why f has no inverse as is.
(b) From here on, assume that the domain of f has been reduced to [1, ∞). Then the
function f : [1, ∞) → [−1, ∞) is injective and its range is [−1, ∞) so that it has an
inverse f −1 : [−1, ∞) → [1, ∞). Sketch the graph of f together with the graph of
f −1 .
(c) Find a formula for the function f −1 .
(d) Verify that f and f −1 satisfy the inverse function identities (i) and (ii) (see 2.5).
Solution:
(a) Since f (2) = 0 = f (0), the two inputs 2 and 0 yield the same output 0, so f is not
injective (and therefore it has no inverse).
(b) The graph of f is a parabola that opens up (after expanding, the coefficient of the x2
term is positive). The vertex is (1, −1) (the smallest the expression can be is −1 and
this occurs when x is 1). Reducing the domain to [1, ∞) retains only the right-hand
side of the parabola. The graph of f −1 is obtained by reflecting through the 45◦ line
y = x:
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(c) f −1 (y) is the x for which f (x) = y, so we solve this last equation for x:
f (x) = y ⇒ (x − 1)2 − 1 = y
p
⇒ x−1=± y+1
p
⇒ x−1= y+1
p
⇒ x = y + 1 + 1,
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(1)
where, at (1), the positive
√ sign was chosen since x is in [1, ∞), implying that x−1 ≥ 0.
Therefore, f −1 (y) =√ y + 1 + 1, or, switching to the more customary input variable
x, we get f −1 (x) = x + 1 + 1.
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(d) For x ∈ [1, ∞), we have
f −1 (f (x)) =
p
f (x) + 1 + 1 =
p
((x − 1)2 − 1) + 1 + 1
= |x − 1| + 1 = (x − 1) + 1 = x,
where the penultimate equality uses that x ≥ 1. This gives (i).
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For x ∈ [−1, ∞), we have
√
f (f −1 (x)) = (f −1 (x) − 1)2 − 1 = (( x + 1 + 1) − 1)2 − 1 = (x + 1) − 1 = x,
giving (ii).
3–1
Simplify the following expressions:
(a) 16−3/4
(b) 81/6 83/2
(c) log3
1
27
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4 ln 3
(d) e
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Solution:
1
1
(a) 16−3/4 = √
=
8
( 4 16)3
√
1
3
10
5
(b) 81/6 83/2 = 8 6 + 2 = 8 6 = 8 3 = ( 3 8)5 = 32
(c) log3
1
27
= −3, since 3−3 =
1
27
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4
(d) e4 ln 3 = eln 3 = 34 = 81, using law of logarithms (iii) and then inverse property of
logarithm (ii).
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3–2
Sketch the graph of the equation y = 2 − ex+1 in steps:
(a) Sketch the graph of y = ex+1 . (Hint: The graph is a shifted version of the graph of
y = ex shown in 3.2.)
(b) Sketch the graph of y = −ex+1 . (Hint: The graph is a reflected version of the graph
of y = ex+1 .)
(c) Sketch the graph of y = 2 − ex+1 . (Hint: The graph is a shifted version of the graph
of y = −ex+1 .)
Solution:
(a)
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(b)
(c)
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3–3
2
(a) Solve e1−x = 2 for x.
(b) Solve log5 (x − 4) + log5 x = 1 for x.
Solution:
(a)
2
e1−x = 2
2
ln e1−x = ln 2
1 − x2 = ln 2
(inverse prop. of log. (i))
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2
x = 1 − ln 2
√
x = ± 1 − ln 2.
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(b)
log5 (x − 4) + log5 x = 1
log5 (x − 4)x = 1
log5 (x−4)x
5
(law of log. (i))
1
=5
(x − 4)x = 5
2
x − 4x − 5 = 0
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(inverse prop. of log. (ii))
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(x − 5)(x + 1) = 0
x−5=0
or x + 1 = 0
x = 5, −1.
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Since log5 (−1) is undefined, the solution x = −1 is extraneous and is therefore
omitted. (In the solution, we applied the law of logarithms (i), which is not valid
unless the factors are both positive, so this is how the extraneous solution entered
in.) The solution is x = 5.
3–4
Show that any logarithm can be expressed in terms of natural logarithms by establishing
the formula
ln x
loga x =
.
ln a
Hint: Write y = loga x in the corresponding exponential form, apply ln to both sides, and
solve for y.
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Solution:
y = loga x
ay = x
ln ay = ln x
y ln a = ln x
ln x
y=
ln a
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(law of logs. (iii))
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3–5
Suppose that the expression 21/3 has not yet been defined. Give an argument similar to
those given
at the end of Section 3.1 to show that 21/3 is forced by the laws of exponents
√
3
to equal 2.
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Solution:
Using part (ii) in the laws of exponents, we have
1
(21/3 )3 = 2( 3 )(3) = 21 = 2.
This says that 21/3 is a number that when cubed equals 2, which is to say 21/3 =
4–1
√
3
2.
Find the following without using a calculator. (Sketch the corresponding point P on the
unit circle as well as the triangle you use to work out the answer, if appropriate):
(a) cos 135◦
(b) sin(4π/3)
(c) sin(−3π/2)
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(d) tan(π/6)
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Solution:
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√
(a) cos 135◦ = −1/ 2
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√
(b) sin(4π/3) = − 3/2
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(c) sin(−3π/2) = 1
triangle 30-60-90, we can use the
(d) Since π/6 = 30◦ is acute and an angle of the known
√
formula at the end of 4.2: tan π/6 = o/a = 1/ 3.
4–2
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Sketch the graph of the equation y = 1 − sin(θ − π/4).
Hint: Use the method of Exercise 3 – 2.
Solution:
The graph is determined in steps:
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