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
Solution to (b): Is x a function of y? Yes.
Take the equation 2x - 5y3 = 1 and try to write x in terms of y:
These calculations justifies the conclusion. Each value of
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
I need to find the x-coordinate(s) of the point(s) of intersection between
the line y = mx and the circle x2 + y2 = 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 x2 + y2 = 1.
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 x2 + y2 = 1 and any line of the form y = mx, we
determine the value of m.
Just as in part (a) we have x2 +(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
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
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
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
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.
Domains Equal? We must have x+1 ≥0 and x ≥0 for the
in the two functions to be real numbers. The next question, is whether
the denominator of f canever be zero ; indeed,
We have argued that if
then 1 = 0; therefore, we
for any x ∈ R.
The domains of these two functions requires x + 1 ≥0 and x ≥0;
Pointwise Equal? Suppose x ∈ Dom(f) = Dom(g), then
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
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,
Exercise Notes: However, we can say
and that for
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).
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
How do you calculate the values of F?
4.6. At the first level f can be broken down into
At a second level:
Each (much simpler) functions are the ones used to "build"
f - through a combination of scalar multiplication, addition,
subtraction and multiplication of functions. Exercise 4.6.
4.7. define f(x) = 2x3 - 1, and g(x) =
sin(x), then F = f5, and
G = g2.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,
What is the natural domain of definition of F? Is there
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.
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