Download 1.7 #6 Meagan

Meagan Holmes
Foundations of Higher Mathematics
“Special Problem” Section 1.7
6. Prove that
(a)
every point on the line ! = ! − ! is outside the circle with radius 4 and
center (3,1).
*Proof by contradiction*
Assume: the line ! = ! − ! and the circle intersect at some point
! =6−!
! + 3 ! + ! − 1 ! = 16
!
! + 3 ! + 6 − ! − 1 = 16
! + 3 ! + 5 − ! ! = 16
! ! + 6! + 9 + 25 − 10! + ! ! = 16
2! ! − 4! + 34 = 16
! ! − 2! + 9 = 0
Equation of the circle
Substitute 6 − ! for !
Simplify
Use Quadratic Equation to find solutions for !
−(−2) ±
−2 ! − 4 1 9
2(1)
Contradiction
=
2 ± 4 − 28
2
If you assume that the line ! = ! − ! and the circle intersect at some point, the
solution would have to be a real number. Since this solution is an imaginary
number, as the above equation shows, this poses a contradiction. Therefore the
original statement is true.
(b)
Prove that there exists a three-digit natural number less than 400 with
distinct digits, such that the sum of the digits is 17 and the product of the
digits is 108.
*Direct proof*
a
b
c
1-3 0-9 0-9
a, b, c ∈ ! + ! + ! = 17
! ∗ ! ∗ ! = 108
Since the three-digit number must be
< 400, !
can only be numbers 1, 2 or 3. ! and ! can be any
0≤
≤ 9, each distinct.
!=1
!=2
1 + ! + ! = 17
1 + 2 + 14 = 17
1 + 3 + 13 = 17
1 + 4 + 12 = 17
1 + 5 + 11 = 17
1 + 6 + 10 = 17
1 + 7 + 9 = 17
1 ∗ 2 ∗ 14 = 28
1 ∗ 3 ∗ 13 = 39
1 ∗ 4 ∗ 12 = 48
1 ∗ 5 ∗ 11 = 55
1 ∗ 6 ∗ 10 = 60
1 ∗ 7 ∗ 9 = 63
Substitute 1 for !
2 + ! + ! = 17
2 + 3 + 12 = 17
2 + 4 + 11 = 17
2 + 5 + 10 = 17
2 + 6 + 9 = 17
2 ∗ 3 ∗ 12 = 72
2 ∗ 4 ∗ 11 = 88
2 ∗ 5 ∗ 10 = 100
2 ∗ 6 ∗ 9 = 108
Substitute 2 for !
By trial and error, we arrive to the conclusion that there exists at least two threedigit natural numbers that satisfy the given statement: 269 & 296.
(c)
Use the Extreme Value Theorem to prove that if ! does not have a
maximum value on the interval [5,7], then ! is not differentiable on (5,7).
*Direct proof using contrapositive*
What we know:
• Not all continuous functions are differentiable, but all differentiable
functions MUST BE continuous.
• Extreme Value Theorem: If a function, !, is continuous on [a,b], then ! has
a maximum and minimum value on [a,b].
• The contrapositive of an implication is equivalent to its original statement
Contrapositive: If ! is differentiable on (5,7), then f has a maximum value on the
interval [5,7].
!→!
If a function, f, is differentiable on (5,7), then it must be continuous on [5,7].
P
!→!
∴!→!
Q
If a f is continuous on [5,7], then it has a maximum value on (5,7).
Q
R
Therefore, if f is differentiable on (5,7), then it has a maximum value on (5,7).
P
R
The contrapositive is true, which proves that the original statement is also true.
(d)
Use Rolle’s Theorem to show that !! + !" − ! = ! does not have more than
one real solution.
*Proof by contradiction*
What we know:
• Rolle’s Theorem: A differentiable function, !, which attains equal
values at two distinct points must have a point somewhere between
them where the first derivative (the slope of the tangent line to the
graph of the function) is zero.
•
Cubic functions can graph two ways:
•
Meaning, the cubic function has either one or three solutions (crosses
the x-axis at one or three points).
Assume:
• the cubic function ! ! = ! ! + 6! − 1 has more than one real solution.
• there exists two points on the graph where the slope of the tangent line
is zero.
! ! = ! ! + 6! − 1
! ! ! = 3! ! + 6
3! ! + 6 = 0
Equation of function
First derivative of function
Set derivative equal to 0
3! ! = −6
! ! = −2
Contradiction
Because ! is not for ! ! = −2, we can conclude that there does not exist a
point on the graph ! ! = ! ! + 6! − 1 where the slope of the tangent line equals
zero, and therefore ! ! does not have more than one real solution. This
contradiction proves the original statement to be true.