Download Math Assignment Solutions: One-to-One Functions, Trigonometry, and Logarithms and more Assignments Calculus in PDF only on Docsity!
MT100.22 ASSIGNMENT TWO—SOLUTIONS
1.4.50 Show that
sin^2 (x) cos^2 (x) =
1 − cos(4x) 8 Solution. Note that
sin^2 (x) =
1 − cos(2x) 2
and cos^2 (x) =
1 + cos(2x) 2
Multiplying these together gives
sin^2 (x) cos^2 (x) =
1 − cos^2 (2x) 4
Using the double angle formula for cosine, we see cos(4x) = 2 cos^2 (2x) − 1. Solving for cos^2 (2x) and substituting, we obtain
sin^2 (x) cos^2 (x) =
( (^) 1+cos(4x) 2
2 − cos(4x) − 1 8
=
1 − cos(4x) 8
1.5.10 Suppose f (x) = (^) x+1^1. Find a domain on which f is one-to-one and a formula for the inverse
of f restricted to this domain. Sketch the graphs of f and f −^1. Solution. The domain of f is all real numbers except for x = −1. Let’s show that f is one-to-one on that whole domain by showing that for x 1 and x 2 in that domain, if f (x 1 ) = f (x 2 ), then it must be the case that x 1 = x 2. Indeed, if f (x 1 ) = f (x 2 ) with x 1 6 = −1 and x 2 6 = −1 we have that 1 x 1 + 1
x 2 + 1
⇒ x 1 + 1 = x 2 = 1 ⇒ x 1 = x 2
Therefore, f is one-to-one. To find a formula for the inverse, swap x and y and solve for y:
x =
y + 1
⇒ y + 1 =
x
⇒ y =
x
1.5.22 Show that f (x) = x+x−^1 is one-to-one on [1, ∞) and find a formula for f −^1 on this domain. What is the domain of f −^1? Solution. To show that f is one-to-one, we may proceed in a couple of ways, such as: (1) Show that f (x 1 ) = f (x 2 ) implies that x 1 = x 2. Suppose that f (x 1 ) = f (x 2 ). Thus
x 1 +
x 1
= x 2 +
x 2
x^21 + 1 x 1
x^22 + 1 x 2 which implies that x^21 x 2 + x 2 − x^22 x 1 − x 1 x 1 x 2
As x 1 , x 2 ≥ 1 (they are both in the domain, which is [1, ∞)) we may multiply through by x 1 x 2 to see that x^21 x 2 + x 2 − x^22 x 1 − x 1 = 0 Factoring, we see that x 1 (x 1 x 2 − 1) − x 2 (x 1 x 2 − 1) = 0 and thus (x 1 − x 2 )(x 1 x 2 − 1) = 0. To solve, we have either (x 1 x 2 − 1) = 0 or (x 1 − x 2 ) = 0. The only way that the first is possible for x 1 , x 2 ∈ [1, ∞) is if x 1 = x 2 = 1. The second says that x 1 = x 2. This shows that f must be one-to-one on the domain [1, ∞). (2) Show that f is strictly increasing on [1, ∞). To do this, suppose that 1 ≤ x < y. Then 1 < xy, and so (^) xy^1 < 1. Therefore 1 x
y
y − x xy
< y − x.
It follows that f (x) = x +
x
< y +
y
= f (y)
and so f is strictly increasing on [1, ∞), and therefore f is one-to-one on that domain. To find a formula for f −^1 , swap x and y and solve for y:
x = y +
y
⇒ xy = y^2 + 1.
Note that we may multiply through by y as y ≥ 1 and in particular is not zero. Now then, we must solve y^2 − xy + 1 = 0, and we may use the quadratic formula to do so: y =
x ±
x^2 − 4 2
To determine the inverse, we must choose either “+” or “−” in the above. Indeed, we cannot choose, it is determined for us! For example, f (2) = 52 , and therefore f −^1
2
must be the case. This demands that the inverse must be given by
f −^1 (x) =
x +
x^2 − 4 2
The domain of f −^1 is [2, ∞).
1.5.30 Compute sin−^1
sin 43 π
without using a calculator. Solution. Note that sin 43 π = −
√ 3 2.^ By the defintion of arcsine, we have that the value must be in the fourth quadrant, and so
sin−^1
sin
4 π 3
−π 3
1.5.40 Simplify cos(tan−^1 (x)) by referring to an appropriate triangle or trigonometric identity. Solution. Design a triangle such that tan(θ) = x. Then tan−^1 (x) = θ. Using this triangle, we may compute cos(θ). That is
cos
tan−^1 (x)
= cos(θ) =
x^2 + 1
