**Solutions to Exercises**

**2.41. **Solution of (a). Is y a function of x? **Yes.**

Indeed, take the given equation
and solve, if possible,

for y. We get

Thus, for each value of x there is only one corresponding
value of y.

We could name this function as
It is easy to see

that the **natural domain** is Dom(f) = R.

**Important Fact.** (The Existence and Uniqueness of Odd Roots) The

pivotal fact used in this calculation is that any real number has a

unique real cube root . More generally, if n ∈ N is an odd natural

number and z ∈ R is any real number, then
exists as a real

number and is unique. What this means in terms of solving equations

is

n odd

Solution to (b): Is x a function of y?** Yes.**

Take the equation 2x - 5y^{3} = 1 and try to write x in terms of y:

These calculations justifies the conclusion. Each value of
y determines

only one corresponding value of x. Let's name this function

It is easy to see that Dom(g) = R is the **natural domain**
of g.

**Exercise 2.41.**

**2.42.** (a) Yes. (b) No.

Solution to (a): Is s a function of t?

Try to solve for s in terms of t. (**Why**?)

That was simple. Thus, each value of t yields only one
value of s; s is

indeed a function of t. We can use the notation

Solution to (b): Is t a function of s?

Try to solve for t in terms of s. (**Why**?)

This shows that for each value of s (for which the
radicand is nonnegative)

there is two value of t. Thus, t is not a function of s.

Example Notes. In line **(A-2)** we used the infamous quadratic formula.

The solution to the equation

is

This is why I set up the equation in line **(A-1).
Exercise 2.42.**

**2.43. **(a) No. (b) No.

Solution to (a): Is x a function of m?

I need to find the x-coordinate(s) of the point(s) of intersection between

the line y = mx and the circle x^{2} + y^{2} = 1.

At any point of intersection ( x, y ), the two variables satisfy both

equations simultaneously : If ( x, y ) is a point of intersection then

y = mx and x^{2} + y^{2} = 1.

Thus,

Now, solving this last equation for x gives the
x-coordinate(s) of the

point(s) of intersection.

Equation **(A-3)** indicates to me that x is not a
function of m: Each

value of m yields two values of x.

Solution to (b): Is m a function of x?

Given that x is the x-coordinate of the point(s) of intersection between

the circle x^{2} + y^{2} = 1 and any line of the form y = mx, we
must

determine the value of m.

Just as in part (a) we have x^{2} +(mx)^{2}
= 1, but now we want to solve

for m in terms of x (**Why**?)

This shows that form each x, x ≠ 0, -1≤ x ≤1, there
corresponds

two values of m. This means that m is not a function of x.

Example Notes: I assume you have drawn a picture of the described

situation. The two algebraic solutions were more difficult than merely

looking at the picture of the situation and reaching the proper conclusions

based on your understanding of the concepts. Equally well, it

is obvious, geometrically, why x ≠ 0 in **(A-4)**: When x is zero, there

is no corresponding m at all; for in that case the corresponding line

is vertical, the slope of a vertical line is undefined. **Exercise 2.43.**

**2.44.** I cannot justify it in your own mind, only in
mine.

**Exercise 2.44.**

**2.45. **Perhaps one could call this the horizontal
line test?

A curve C in the xy -plane defines x as a function
of y if it
is true that every horizontal line intersects the curve at no
more than one point. |

If this be true, then for each y there corresponds at most
one x -

this is descriptive of the concept of x is a function of y. (Note: if for a

given y the horizontal line at altitude y does not intersect the curve,

this means that y does not belong to the domain of the function.)

**Exercise 2.45.**

**2.46. **Simply restate the **Function Line Test **
using these different

letters:

Solution to (a): C defines s as a function of t provided every line

perpendicular to the t-axis intersects the curve at no more than one

point.

Solution to (b): C defines t as a function of s provided every line

perpendicular to the s-axis intersects the curve at no more than one

point .**Exercise 2.46.**

**2.47. **Let C be a curve in the pq-plane.

Under what conditions can we assert that p is a function of q? Review

the definition of function in your head, and answer this question

without error. Passing Score. 1 out of 1.

(a) Every line perpendicular to

the p-axis intersects the curve

at no more than one point.

(b) Every line perpendicular to

the q-axis intersects the curve

at no more than one point.

**Exercise 2.47.**

**4.1.** Yes.

**Domains Equal?** We must have x+1 ≥0 and x ≥0 for the
radicals

in the two functions to be real numbers. The next question, is whether

the denominator of f can ever be zero ; indeed,

We have argued that if
then 1 = 0; therefore, we

conclude
for any x ∈ R.

The domains of these two functions requires x + 1 ≥0 and x ≥0;

therefore,

**Pointwise Equal?** Suppose x ∈ Dom(f) = Dom(g), then

Thus,

Conclusion: Yes indeed, f = g.

Example Notes: You did remember the trick of multiplying by the

conjugate, didn't you? It is also important to note that the quantity

is never equal to zero; therefore,
multiplying the numerator

and denominator by this quantity is equivalent to multiplying by

one, no matter what the value of x. **Exercise 4.1.**

**4.2.** No. Recall,

**Domains Equal?** I think we can take a useful shortcut. Note
that

since x = 0 makes the denominator equal to
zero (and the

numerator too ). But 0 ∈ Dom(g) since we can calculate

for the case of x = 0; indeed, g(0) = 0. Thus, we have argued that

and 0 ∈ Dom(g). Therefore,

and so,

All done!

Exercise Notes: However, we can say
and that for

all
we have f(x) = g(x). In this case, we say
that g is an

extension of f. **Exercise 4.2.**

**4.3. **Look at the calculating formula. The concept
is to add the values

of f and g together. So given an x, we need to have defined the value

f(x). This implies x∈ Dom(f). Similarly, we need to add f(x) to g(x)

-g(x) needs to be a defined quantity; therefore, x ∈ Dom(g) as well.

We have argued that in order to carry out the concept of summing

two functions together, we must choose an x in both Dom(f) and in

Dom(g). Thus, 'Nuff said. **Exercise 4.3.**

**4.4.** Yes. The domain of F is void (empty).

and ,

Thus,

The "function" F has no domain, hence, F remains
undefined.

**Exercise 4.4.**

**4.5.** This is a simple generalization of the
sum/difference of two

functions.

How do you calculate the values of F?

**Exercise 4.5.**

**4.6.** At the first level f can be broken down into
two pieces.

and

At a second level:

Each (much simpler) functions are the ones used to "build"
the functions

f - through a combination of scalar multiplication, addition,

subtraction and multiplication of functions. **Exercise 4.6.**

**4.7.** define f(x) = 2x^{3} - 1, and g(x) =
sin(x), then F = f^{5}, and

G = g^{2}.**Exercise 4.7.**

**4.8.** Three! But is it the three you had in mind?
define f(x) = x,

g(x) = sin(x), and h(x) = 1. Then,

or,

What is the natural domain of definition of F? Is there
another way

of writing F?**Exercise 4.8.**

**4.9.** We say that f < g over the set A provided

Geometrically, this means that the graph of f is always below the

graph of g when plotting these functions over the set A.

To be true to our mathematical roots we should realize what about

the set A?

Part of successfully answering this question is being able
to read and

understand the proposed questions. Then after some thought, respond

correctly with error. How did you do?** Exercise 4.9.**

**5.1.** I'll leave the details to you. But I will tell
you,

Verify! **Exercise 5.1.**

**5.2. **A challenge to our perspicacity, without
doubt.

Calculation of W o M. By convention, we take as the independent

variable symbol, the symbol used by the inner-most function for its

independent variable. Thus

But, of course, the independent variable is a **dummy**,
so it really didn't

matter what letter we used. Thus,

all define exactly the same function. The use of u was
just a little

more convenient to use than the other variables.

Calculation of M o W. By convention, we take as the
independent

variable symbol, the symbol used by the inner-most function for its

independent variable. Thus

Ditto the comments made above. **Exercise 5.2.**

**5.3.** The height of triviality:
I hope you got it.

**Exercise 5.3.**

**5.4.** Let
, then

y = tan(u), where
.

**Exercise 5.4.**

**7.1.** This is a polynomial in t of degree 5 with
rational coefficients.

**Exercise 7.1.**

**7.2.** First "proof by example." This is not a proof,
but it is frequently

used to get insight into how to formally prove an assertion. Take as

an example:

Get a common denominator for the coefficients - that would
be 6.

Thus,

Now, factor out 1/6.

As you can see, the original polynomial is written as a
rational scalar

multiple (that is, a scalar multiple that is a rational number) times a

polynomial with integer coefficients.

The general proof is an abstract manifestation of this example.

**Exercise 7.2.**