Math 1210
Quiz 7
March 7th, 2014
This is a take-home quiz and it is due on TUESDAY, MARCH 18th. It is a
bit longer than usual. The fact that the instructions are long is not meant to
put you off: they are meant to be a guide and the idea is that you use the
textbook and your notes in order to learn as much as possible by solving the
exercises. The value of every question is indicated at the beginning of it.
Please turn in this sheet only and feel free to email me during the break if
you have questions.
Name:
UID:
1. (15 points) Consider the function f (x) = x3 + 6x2 + 9x + 1. In this exercise you we will
use the intermediate value theorem for derivatives to show that the equation f (x) = 0
has a unique solution inside of the interval [−1, 1]
1. (3 points) State the intermediate value theorem (for continuous functions) that we
studied in section 1.6 and use it to show that the equation f (x) = 0 has at least
one solution inside of [−1, 1].
Solution: The IVT for continuous functions states that if f is a continuous function
defined over a closed interval [a, b], then for every value K between f (a) and f (b)
there exists some c inside [a, b] such that f (c) = K. In particular, if the values of f
at the endpoints have opposite signs, there exists some c in [a, b] such that f (c) = 0.
In our case, f (−1) = −1 + 6 − 9 + 1 = −3 < 0 and f (1) = 1 + 6 + 9 + 1 = 17 > 0
have opposite signs, so there exists some c inside [−1, 1] such that f (c) = 0.
2. To show that there is only one, we argue by contradiction.
• (2 points) State the intermediate value theorem (IVT) for derivatives (section
3.6).
Solution: The IVT for derivatives states that if f is a function which is continuous over the closed interval [a, b] and differentiable over the open interval
(a)
(namely,
(a, b), then there exists some c inside (a, b) such that f 0 (c) = f (b)−f
a−b
the tangent line to the curve y = f (x) at c is parallel to the line joining the
points (a, f (a)) and (b, f (b))).
• (2 points) Suppose that the equation f (x) = 0 has 2 solutions inside the interval
[−1, 1]. What does the IVT say about f 0 (x)in this case?
Solution: If there are two x-values c1 , c2 inside [−1, 1] such that f (c1 ) =
f (c2 ) = 0, since f is differentiable the IVT for derivatives guarantees that there
exists some c inside [c1 , c2 ] (in particular, c lies in [−1, 1]) such that f 0 (c) = 0.
• (2 points) Compute f 0 (x) and show that that the above conclusion cannot hold.
Solution: The derivative of f is given by f 0 (x) = 3x2 + 12x + 9 = 3(x2 +
4x + 3) = 3(x + 1)(x + 3) and it is only zero at −1 and −3. None of these
x-values lies inside of the interval (−1, 1) as we concluded previously so there
can’t possibly be more than one x-value inside (−1, 1) at which f vanishes.
3. (6 points) Argue in the same way that the equation f (x) = 0 has a unique solution
inside the interval [−3, −1].
Solution: Since f is continuous over [−3, 1], f (−3) = −27 + 54 − 27 + 1 = 55 > 0
and f (−1) = −3 < 0, the IVT for continuous functions implies that there exists at
least one x-value c inside of (−3, −1) such that f (c) = 0. If there were at least 2
such x-values, since f is differentiable over [−3, −1], the IVT for derivatives would
imply the existence of an x-value c inside (−3, −1) such that f 0 (c) = 0. But we
just saw that f 0 only vanishes at −1 and −3, so f can have one single zero inside
(−3, −1).
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2. (15 points) Consider again the function f (x) = x3 + 6x2 + 9x + 1.
1. (1 point) What is the domain of f ?
Solution: f is a polynomial, so it domain is (−∞, +∞).
2. (2 point) Compute the limits limx→+∞ f (x) and limx→−∞ f (x).
Solution: limx→−∞ f (x) = −∞ and limx→+∞ f (x) = +∞.
3. (4 points) Over which intervals is f increasing (resp. decreasing)? Namely, for
which x is f 0 (x) positive (resp. negative)?
Solution: The derivative of f is given by f 0 (x) = 3x2 + 12x + 9 = 3(x2 + 4x + 3) =
3(x + 1)(x + 3) and it is zero at −1 and −3
(i) If x < −3, then f 0 (x) > 0, so f is increasing over (−∞, −3).
(ii) If −3 < x < −1, then f 0 (x) < 0, so f is decreasing over (−3, −1).
(iii) If x > −1, ten f 0 (x) > 0, so f is increasing over (−1, ∞).
4. (2 points) What are, if any, the local extreme values of f ?
Solution: By the first derivative criterion, f has a local maximum value at x = −3
(f goes from being increasing to being decreasing) and a local minimum value at
x = −1 (f foes from being decreasing to being increasing). Since f has no singular
points, these are all the local extreme values of f .
5. (4 points) Over which intervals is f concave up (resp. down)? Namely, for which x
is f 00 (x) positive (resp. negative)?
Solution: The second derivative of f is given by f 00 (x) = 6(x + 2) and it is zero at
−2.
(i) If x < −2, then f 00 (x) < 0, so f is concave down over (−∞, −2).
(ii) If x > −2, ten f 00 (x) > 0, so f is concave up over (−2, +∞).
6. (2 points) What are, if any, the inflection points of f ?
Solution: f has an inflection point with x-value −2 (f goes from being concave
down to being concave up). Since f has no singularities, there are no other inflection
points.
7. (4 points) Sketch the graph of f .
Solution: We start by computing the values of f at representative points (singular
points, local extreme points and inflection points). We have seen that f has a local
maximum at x = −3) with value f (−3) = 1 and that f has a local minimum at
x = −1 with value f (−1) = −3. There is also an inflection point with x-value
x = −2 and f (−2) = −1.
8. Check that your computations and your graph are consistent with the conclusions
of Exercise 1.
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Figure 1: Graph of the function f (x) = x3 + 6x2 + 9x + 1.
3. (15 points) Consider again the function f (x) = x3 + 6x2 + 9x + 1. You have shown
in Exercise 1 that the equation f (x) = 0 has a (unique) solution inside of [−1, 1]. Use
Newton’s method to find an approximation of it which is accurate to 5 decimal places.
1. (6 points) Recall that, starting from an appropriate initial approximation x1 , Newton’s method provides a sequence of points x2 , x3 , . . . that converges towards the
solution of the equation. Starting with x1 = 2, fill in the following table until
|xn − xn−1 | < 0.00001 (6 steps will be enough).
n
1
2
3
4
5
6
7
n−1 )
xn = xn−1 − ff0(x
(xn−1 )
x1 = 2
x2 = 0.866666667
x3 = 0.222071702
x4 = −0.057751015
x5 = −0.117876028
x6 = −0.120609208
x7 = −0.120614758
|xn − xn−1 |
−
|x2 − x1 | = 1.133333333
|x3 − x2 | = 0.644594964
|x4 − x3 | = 0.279822717
|x5 − x4 | = 0.060125013
|x6 − x5 | = 0.00273318
|x7 − x6 | = 0.00000555
We thus conclude that x7 = −0.120614758 is an approximation of the solution to the
equation f (x) = 0 which is accurate to 5 decimal places (to check that x7 is indeed
close to the real solution of the equation f (x) = 0, note that f (x7 ) = 1.7366910−10 ).
2. (1 point) Why was it a good idea to use x1 = 2 as an initial approximation?
Solution: Newton’s method uses derivatives in order to approximate zeros of functions. Therefore, one must take as an initial approximation an x-value which is
”close enough” to the real solution, in the sense that the behavior of the function
f should not change in between the two. Looking at the graph of the function that
we sketched in exercise 2, we see that any x-value in the interval (−1, +∞) would
lead to a correct approximation (whereas starting from an x-value located in the
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interval (−3, −1), for instance, would lead to the wrong solution, as we will see in
the next part).
3. (6 points) Now take x1 = −1.5 as your initial approximation and run Newton’s
method again (fill in the table below). What is the output of the algorithm in this
case? What feature of f do you think explains this (look at the graph for f (x) that
you sketched in Exercise 2).
n
1
2
3
4
5
6
n−1 )
xn = xn−1 − ff0(x
(xn−1 )
x1 = −1.5
x2 = −2.555555556
x3 = −2.316798942
x4 = −2.346958322
x5 = −2.34729631
x5 = −2.347296355
|xn − xn−1 |
−
|x2 − x1 | = 1.055555556
|x3 − x2 | = 0.238756614
|x4 − x3 | = 0.030159381
|x5 − x4 | = 0.000337988
|x6 − x5 | = 4.50859 · 10−8
In this case we see that x6 = −2.347296355 approximates a solution to the equation
f (x) = 0 which is accurate to 7 decimal places, but this is not the solution we were
looking for!. Remember that the goal was to find an approximation to the solution
of the equation f (x) = 0 that is located inside the interval (−1, 1). Since we
picked x1 = −1.5 as an initial approximation, and this x-value is smaller than
the local minimum at x = −1, we see that the sequence of points that Newton’s
method produces (by intersecting the tangent lines with the x-axis) converges to
the solution that is located inside (−3, −1).
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