Solutions to activities

Solution to activity 2.1

Using the triangles in Figure 2.5 and the definition of the tangent function, it should be clear that

tanπ 6 = 1

√3, tanπ

4 = 1 and tanπ 3 =√

3.

Indeed, using the fact that

angle in radians = 2π

360 × angle in degrees,

we can see that an angle of π/6, π/4 or π/3 radians corresponds to an angle of 30, 45 or 60 degrees respectively.

Solution to activity 2.2

In this case, the unit circle method gives us the situation illustrated in Figure 2.16 and so the angle in the triangle would be π/3 (i.e. π−^2π₃ = ^π₃ as the angle subtended by a

2

2.2. Solutions to activities

straight line — in this case the x-axis — is π) giving x a magnitude of 1/2 and y a magnitude of √

3/2 whereas their signs would be negative for x (as x < 0) and positive for y (as y > 0). Thus we see that

sin2π 3 =

√3

2 and cos2π

3 =−1 2, using the unit circle method.

O x

y 1 (x, y)

2π/3

Figure 2.16:For Activity 2.2, we find sin θ and cos θ when θ = 2π/3 by considering a unit circle.

Solution to activity 2.3

Using the unit circle in Figure 2.17(a), it should be clear that sin 0 = 0 and cos 0 = 1,

whereas using the unit circle in Figure 2.17(b), it should be clear that sinπ

2 = 1 and cosπ 2 = 0.

Then, using similar reasoning, we should be able to deduce that

θ π 3π

2 2π

sin θ 0 −1 0

cos θ −1 0 1

are the other values of the functions sin θ and cos θ that we seek.

Solution to activity 2.4

From the definition of cot θ, we have cot θ = 1

tan θ = 1 sin θ cos θ

= cos θ sin θ,

as we know that tan θ = sin θ/ cos θ. This function is defined as long as θ 6= nπ for n ∈ Z since, at these values of θ, we have tan θ = 0 or, equivalently, sin θ = 0.

Solution to activity 2.5

Given the functions f : R→ R and g : R → R where f(x) = x²+ 1 and g(x) = 2^x, we see that

2

O x

y

1 O

x y

1

(a) (b)

Figure 2.17: For Activity 2.3, we find sin θ and cos θ by considering a unit circle when (a) θ = 0 and (b) θ = π/2.

g◦ f is the function where

(g◦ f)(x) = g(f(x)) = g(x²+ 1) = 2^x2+1, where (g◦ f) : R → R.

f ◦ g is the function where

(f ◦ g)(x) = f(g(x)) = f(2^x) = (2^x)²+ 1 = 2^2x+ 1, and (f ◦ g) : R → R.

Indeed, observe that as 2^x2+1 6= 2^2x+ 1, these are certainly not the same function.

Solution to activity 2.6

If we have f (x) = x³ and g(x) = x²+ 5, then the function (x²+ 5)³ can be written as (x²+ 5)³ = f (x² + 5) = f (g(x)) = (f◦ g)(x),

i.e. it is the composition we get from applying f after g or, in terms of x, it is the composition of the function x³ after the function x²+ 5.

Solution to activity 2.7

By considering the graphs of the functions f (x) = x + 2 and f⁻¹(x) = x− 2 as

illustrated in Figure 2.18, we see that the latter is indeed the reflection in the line y = x of the former. Alternatively, we can see that if y = x + 2, a reflection in the line y = x just means replacing all points (x, y) that satisfy this equation with points given by (y, x) to get the new equation x = y + 2. But, of course, this gives y = x− 2 which is what we wanted.

Solution to activity 2.8

When we considered the function f : R→ R in Example 2.5, there were two problems that prevented us from finding an inverse. To counteract these so that we can find the local inverses of this function, we note that:

If we take the co-domain to be the interval [0,∞) so that we have y ≥ 0, then we remove the problem that occurs because y = x² has no solution for y < 0.

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2.2. Solutions to activities

O x

y

y=x+2 y=x

y=x − 2 2

−2

−2 2

Figure 2.18:For Activity 2.7, we see that the graph of f⁻¹(x) = x− 2 is the reflection of the function f (x) = x + 2 about the line y = x.

If we take the two domains given by the intervals (−∞, 0] and [0, ∞) so that we have x ≤ 0 and x ≥ 0 respectively, then we remove the problem that occurs because y = x² has two solutions for x∈ R.

Indeed, this means that if we consider the function

f : [0,∞) → [0, ∞) given by f(x) = x², then we have

y = f (x) =⇒ y = x² =⇒ x =√ y,

as x≥ 0 because x ∈ [0, ∞). Thus, using x = f⁻¹(y), the inverse of this function is f⁻¹(y) = √y or f⁻¹(x) =√

x if we want it in terms of x.

f : (−∞, 0] → [0, ∞) given by f(x) = x², then we have

y = f (x) =⇒ y = x² =⇒ x =−√y,

as x≤ 0 because x ∈ (−∞, 0]. Thus, using x = f⁻¹(y), the inverse of this function is f⁻¹(y) =−√y or f⁻¹(x) =−√

x if we want it in terms of x.

In particular, this means that the local inverses of f : R→ [0, ∞) where f(x) = x² are f⁻¹(x) =√

x when x∈ [0, ∞) and f⁻¹(x) =−√

x when x∈ (−∞, 0].

Solution to activity 2.9

We saw in Activity 2.8 that the function f : [0,∞) → [0, ∞) where f(x) = x² has an inverse given by f⁻¹(x) =√

x. The graphs of these two functions are illustrated in Figure 2.19. In particular, observe that the curve y =√

x is the reflection about the line y = x of the curve y = x² and that all three of these curves intersect at the points (0, 0) and (1, 1).

Solution to activity 2.10

From the graphs of the function f (x) = xⁿ where f : R→ R when n is odd, which we saw in Figure 2.2(a) and (c), it should be clear that the the equation y = f (x) has a unique solution, x, for all y∈ R and so the inverse of this function exists. In particular, we see that

y = xⁿ =⇒ x = y^1/n = √ⁿ y,

gives us this unique solution for any y∈ R provided that n is odd and so we have f⁻¹(y) = √ⁿy as the inverse function or, indeed, f⁻¹(x) = √ⁿ

x if we want it in terms of x.

2

O x

y y=x

y =√ x y=²x

1 1

Figure 2.19: For Activity 2.9, we see that the graph of f⁻¹(x) = √

x is the reflection of the function f (x) = x² about the line y = x.

However, from the graph of the function f (x) = xⁿ where f : R→ R when n is even, which we saw in Figure 2.2(b), it should be clear that when

y < 0, the equation y = f (x) has no solution for x as we know that x∈ R means that y = xⁿ≥ 0 when n is even.

y > 0, the equation y = f (x) has two solutions for x as we know that we can get x =±√ⁿy∈ R when n is even.

As such, we can not find a unique solution, x, for all y ∈ R and so the inverse of this function can not exist.

Solution to activity 2.11

We saw the graph of a function like f : R→ (0, ∞) where f(x) = 2^x in Figure 2.3(b) since we have a = 2 > 1 here. As such, we find that the graphs of the function

f : R→ (0, ∞) where f(x) = 2^x and its inverse, f⁻¹(x) = log₂x where

f⁻¹ : (0,∞) → R, are as illustrated in Figure 2.20. In particular, observe that the curve y = log₂x is the reflection about the line y = x of the curve y = 2^x.

y

1 O 1

x y =log²x

y=x y=^x2

Figure 2.20: For Activity 2.11, we see that the graph of f⁻¹(x) = log₂x is the reflection of the function f (x) = 2^x about the line y = x.

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2.2. Solutions to activities

Solution to activity 2.12

To find the acute angles θ1 and θ2 where θ1 = sin^{−1 1}₂ and θ2 = cos^{−1 1}₂, we use the table of values in Section 2.1.1 to see that

sin θ1 = 1 Activity 2.1 to see that

tan θ3 = 1 = tanπ using the definitions of the reciprocals of our three trigonometric functions, which we saw in Section 2.1.2.

Solution to activity 2.13

Given that f (x) = x³, g(x) = x⁴ and h(x) = 2^x, we use the definitions of the combinations of functions we need from Section 2.1.2, to get

(f · g)(x) = f(x)g(x) = (x³)(x⁴) = x⁷,

where we have used the power laws to simplify our answers. Indeed, observe that for the last function, we can also write 2^4x= (2⁴)^x = 16^x.

Solution to activity 2.14

To derive the laws of logarithms, we note that for the first one, we use the power laws and the given fact to get

a^logax+logay = a^logaxa^logay = xy = a^loga(xy),

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Solution to activity 2.15

We take logarithms to the base b on both sides of the given fact to see that a^logax = x =⇒ log_b a^logax

= log_bx =⇒ (log_ax)(log_ba) = log_bx,

where we have used the third law of logarithms in the last step. Then, dividing through on both sides by log_ba (which is non-zero as a6= 1), we get result from Activity 2.4. Then, again starting with sin²θ + cos²θ = 1, we divide both sides by cos²θ to get

With the given facts, we can use the compound-angle formula for sin(θ + ϕ) to see that sin(θ− ϕ) = sin(θ + (−ϕ)) = sin θ cos(−ϕ) + cos θ sin(−ϕ) = sin θ cos ϕ − cos θ sin ϕ, and the compound-angle formula for cos(θ + ϕ) to see that

cos(θ− ϕ) = cos(θ + (−ϕ)) = cos θ cos(−ϕ) − sin θ sin(−ϕ) = cos θ cos ϕ + sin θ sin ϕ, whereas using the compound-angle formula

cos(θ + ϕ) = cos θ cos ϕ− sin θ sin ϕ,

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2.2. Solutions to activities

with ϕ = θ we get

cos(θ + θ) = cos θ cos θ− sin θ sin θ =⇒ cos(2θ) = cos²θ− sin²θ,

as required. Indeed, since we also have the Pythagorean identity sin²θ + cos²θ = 1, we can write this last double-angle formula as

cos(2θ) = (1− sin²θ)− sin²θ = 1− 2 sin²θ, in terms of sin²θ, or as

cos(2θ) = cos²θ− (1 − cos²θ) = 2 cos²θ− 1, in terms of cos²θ, as required.

Solution to activity 2.19

From the graph in Figure 2.11, we can see that the economically meaningful part of the supply function is q^S : [0,∞) → [1, ∞) where q^S(p) = p + 1 and the economically

meaningful part of the demand function is q^D : [0, 3]→ [0, 3] where q^D(p) = 3− p.

Clearly, both of these functions are invertible as each q in the co-domain gives rise to a unique p in the domain and we find that

q = p + 1 =⇒ p = p^S(q) = q− 1, is the inverse supply function, whereas

q = 3− p =⇒ p = p^D(q) = 3− q, is the inverse demand function.

Solution to activity 2.20 Given that

y = a(x− p)²+ q, we see that

If a > 0, then for any x∈ R,

(x− p)² ≥ 0 =⇒ a(x− p)² ≥ 0 =⇒ a(x− p)²+ q ≥ q,

i.e. for all x ∈ R, y ≥ q and so the smallest value of y occurs when y = q which, in turn, means that we must have x = p. Thus, the turning point of the parabola is a minimum and this occurs at the point (p, q).

If a < 0, then for any x∈ R,

(x− p)² ≥ 0 =⇒ a(x− p)² ≤ 0 =⇒ a(x− p)²+ q ≤ q,

i.e. for all x ∈ R, y ≤ q and so the largest value of y occurs when y = q which, in turn, means that we must have x = p. Thus, the turning point of the parabola is a maximum and this occurs at the point (p, q).

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Solution to activity 2.21

Given that the equation of the hyperbola is y²

9 − x² 4 = 1,

we see that there are no x-intercepts since, setting y = 0, we get

−x²

4 = 1 =⇒ x² =−4,

which has no real solutions, whereas we see that the y-intercepts, which occur when x = 0, are given by

y²

9 = 1 =⇒ y² = 9 =⇒ y =±3.

To find the asymptotes, we write the equation as y²

x² = 9

1 4 + 1

x²

 ,

so that, as x→ ∞, we have 1/x² → 0 and this leaves us with y²

x² = 9

4 =⇒ y =±3 2x,

as the equations of the asymptotes. Putting this information together, we then get the sketch in Figure 2.21.

x y

O 3

−3 y

= −

₃

2

x y =

3 2

x

y²

9

−

^x₄²

= 1

Figure 2.21: For Activity 2.21, a sketch of the hyperbola y² 9 − x²

4 = 1.

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2.2. Exercises

Exercises

Exercise 2.1

Sketch the graph of the function f :{x ∈ R | x 6= −1, 1} → R given by

f (x) = x⁴− 1 x²− 1.

Exercise 2.2

Use the compound-angle formulae to show that

tan(θ± ϕ) = tan θ± tan ϕ 1∓ tan θ tan ϕ, and hence deduce an expression for tan(2θ).

Exercise 2.3

The supply and demand functions for a good are

q^S(p) = p− 4 and q^D(p) = 8− p,

respectively. Sketch the graphs of these functions and find the equilibrium point.

A percentage [of the price] tax of 100r% is imposed. Find the new equilibrium point and, by sketching the graph of the new supply function on your earlier sketch, comment on how the equilibrium point for the market has changed. How much of the tax has been passed onto the consumers? What is the maximum tax, rm, that can be imposed if this market is to continue functioning?

Exercise 2.4

When selling a quantity, q, a firm makes a profit given by π(q) = q²+ 2q + 2,

and the largest quantity it can produce is 10. Sketch the graph of this profit function and deduce the value of q that will yield the greatest profit for this firm.

Explain why the inverse profit function exists and find it.

Exercise 2.5

Sketch the circle and the rectangular hyperbola with equations x²+ y² = 1 and 2xy = 1, respectively. At what points do these two curves intersect?

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In document Calculus (Page 44-54)