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Worksheet #15
Math 221
Instructions: Put the first and last name of everyone in your group at the top of your paper.
Everyone is to do their own worksheet but only one from each group is graded with the score
shared. Be sure to show your work and explain your reasoning.
Some questions on this worksheet reference a computer-generated demonstration (access
it through Moodle) to aid in conceptual understanding of the questions.
1. Understanding what f (x), f 0 (x), and f 00 (x) tell you about the graph of f (x).
(a) Let f (x) =
x2 −4
.
x−c
i. Graphically, how do changes to the parameter c change the graph of f (x)?
Answer: From the portion of the graph we’re shown, we see that changes in
c correspond to changes in the location of the vertical asymptote, except for
c = ±2, when the function appears to just be a line.
ii. Use f (x), f 0 (x), or f 00 (x) to mathematically explain why the graph changed
in the way you observed.
Answer: The function factors to show that (except when c = ±2) it will
have a vertical asymptote when x = c. When c = ±2, the function has a hole
at x = c.
(x − 2)(x + 2)
f (x) =
.
x−c
(b) Let f (x) = x3 − 3cx2 + x + 1.
i. Graphically, how do changes to the parameter c change the graph of f (x)?
Answer: From the portion of the graph we’re shown, we can see that c
changes the location of the inflection points, and the locations of the maximum and minimum values. In order to explain these changes, we will have
to find the first and second derivative.
ii. Use f (x), f 0 (x), or f 00 (x) to mathematically explain why the graph changed
in the way you observed.
Answer: In order to analyze how the location of the maximum and minimum
values change, we need to take the first derivative.
f (x) = x3 − 3cx2 + x + 1 ⇒ f 0 (x) = 3x2 − 6cx + 1,
1
(1)
√
2
which means f has critical points of x = c ± 9c3 −3 . Thus, changing c leads
to changes in the critical points.
In order to explain the inflection points, we need to take another derivative.
f 00 (x) = 6x − 6c = 6(x − c).
(2)
From this, we can see that second derivative will always equal zero when x = c.
For x < c, f 00 (x) < 0 or f (x) is concave down, and for x > c, f 00 (x) > 0 so
f (x) is concave up. This means that x = c serves as an inflection point to the
function and changing c will change the location where concavity changes.
(c) Let f (x) = 2x3 + 3(4 − c)x2 − 24cx + 1.
i. Graphically, how do changes to the parameter c change the graph of f (x)?
Answer: From the portion of the graph we’re shown, we see that changes in
c appear to correspond to changes in the location of the local minimum.
ii. Use f (x), f 0 (x), or f 00 (x) to mathematically explain why the graph changed
in the way you observed.
Answer: Taking the first derivative, we obtain,
f 0 (x) = 6x2 + 6(4 − c)x − 24c = 6(x2 + (4 − c)x − 4c) = 6(x + 4)(x − c).
From this we can see that the critical points to f are x = −4 and x = c.
In this demo, we were only considering −3 ≤ c ≤ 3, so we can assume that
c > −4. Evaluating the first derivative over the intervals (−∞, −4), (−4, c),
and (c, ∞), we have f 0 (x) > 0, f 0 (x) < 0, and f 0 (x) > 0 respectively. This
means that the function experiences a local minimum at x = c for the values
of c we were shown.
2. Curve Sketching: For each of the below functions, answer the following questions.
Determine enough relevant information about functions to provide a good graph.
(a) f (x) = 31 x3 + x2 − 3x. (Two x-intercepts are approximately −4.85 and 1.85.)
i. What does f (x) tell us? The domain is all real numbers since f is a polynomial. Intercepts are (0, 0), (−4.85..., 0), and (1.85..., 0). There is no obvious
symmetry.
ii. Evaluate limits. Polynomials do not have asymptotes. As x → ∞, f (x) → ∞,
and while x → −∞, f (x) → −∞.
iii. What does f 0 (x) tell us? We have
f 0 (x) = x2 + 2x − 3 = (x + 3)(x − 1),
which means the critical points are x = −3 and x = 1. Since f 0 (x) > 0
when x < −3 and when x > 1, we have f (x) is increasing over the intervals
(−∞, −3) and (1, ∞). Similarly, the first derivative lets us conclude that f (x)
is decreasing over the interval (−3, 1). The first derivative test establishes that
f has a local maximum at x = −3 and a local minimum at x = 1.
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iv. What does f 00 (x) tell us? The second derivative is
f 00 (x) = 2x + 2 = 2(x + 1).
From this, we see that the second derivative is negative for x < −1 and
positive for x > −1. This means that f (x) is concave down on the interval
(−∞, −1) and concave up on the interval (−1, ∞) and that x = −1 is an
inflection point.
v. Provide a good sketch of f (x).
(b) f (x) =
x
tan(x)
=
x cos(x)
.
sin(x)
Graph f (x) over the interval [−π, π].
i. What does f (x) tell us? Domain is all values of x excluding all the points
where sin x is equal to zero. Since we’re only considering − ∈≤ x ≤ π,
this means our domain is (−π, 0) and (0, π). Immediately, we note that this
function will have intercepts when cos(x) = 0, which is − π2 and π2 . We might
think that this function also has an intercept when x = 0, but the function
is not defined there (even though the limit exists!).
ii. Evaluate limits. Since we’re only looking at the interval [−π, π], there will
not be horizontal (or slant) asymptotes. Candidates for vertical asymptotes
are when sin(x) = 0, but we need to evaluate the limits to establish whether
or not there is actually an asymptote. Note that
(−π)(−1)
x cos(x)
=
= −∞ &
sin(x)
0−
lim +
x→−π
lim−
x→π
x cos(x)
(π)(−1)
=
= −∞,
sin(x)
0+
which means our function has vertical asymptotes as it approaches ±π. Testing the last candidate, we see that
x cos(x)
x
lim
= lim
cos(x) = 1 · 1 = 1,
x→0 sin(x)
x→0
sin(x)
or f does NOT have a vertical asymptote at x = 0.
iii. What does f 0 (x) tell us?
x
cos(x) − sin(x)
tan(x) − x sec2 (x)
cos(x)
1
f (x) =
=
=
−x 2 .
2
tan (x)
sin(x)
sin(x)
sin (x)
0
3
From this we see that f 0 (x) is not defined at x = 0, so this is a critical point
to the function. For −π < x < 0, f 0 (x) > 0 so f is increasing on that interval.
For 0 < x < π, f 0 (x) < 0, which means f is decreasing on that interval. The
point for which x = 0 is a local maximum of the function.
iv. What does f 00 (x) tell us?
f 00 (x) = −
2
(1 + x cot(x)) < 0 for
sin (x)
2
− π < x < π.
This means that the function is concave down over the interval of consideration.
v. Provide a good sketch of f (x).
√
(c) f (x) = (x − 4) x.
i. What does f (x) tell us? The domain of this function is x ≥ 0. The intercepts
are (0, 0) and (4, 0). No obvious symmetry.
ii. Evaluate limits. There are no vertical asymptotes and lim f (x) = ∞, so the
x→∞
function does not have a horizontal asymptote.
iii. What does f 0 (x) tell us?
f 0 (x) =
√
x+
(x − 4)
3x − 4
√
= √ .
2 x
2 x
Since x = 0 is in the domain of f , but f 0 (0) DNE, x = 0 is one critical point
to the function. The other arises from setting the numerator equal to zero:
x = 43 . We have that f 0 (x) < 0 over (0, 43 ), and f 0 (x) > 0 for 43 < x. This
means the function is decreasing over the interval (0, 34 ) and increasing over
the interval ( 34 , ∞). The first derivative test tells us that the point x = 43 is
a local minimum to the function.
iv. What does f 00 (x) tell us?
3x + 4
f 00 (x) = √ 3 .
4( x)
For x > 0, f 00 (x) > 0, which means the function is concave up over the entire
domain.
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v. Provide a good sketch of f (x).
3. Q: Do polynomials with similar coefficients have similar roots? To answer
this question, we will explore two similar polynomials and their roots. Let
f (x) = x4 − 120x3 + 602x2 + 10000x + 22500
and
g(x) = x4 − 120x3 + 604x2 + 10000x + 22500.
(a) What is the difference between f (x) and g(x)? How much of a difference do you
expect in their graphs?
We do not expect much difference in the graphs, because f (x) = g(x) − 2x2 and
the functions have much larger terms that will dominate that difference.
(b) The function f (x) has four real roots. Approximately, where does f (x) have
roots? Zoom in on the graph to identify all of the roots.
The roots are around x = 114, 13, −3.87, −3.73.
(c) The function g(x) has only two real roots. Approximately, where does g(x) have
roots? Zoom in on the graph to identify all of the roots.
The roots are around x = 114, 13.
(d) Consider the new function h(x) = f (x) + ax2 in which a is a parameter we may
change. When a = 0, h(x) = f (x), but when a = 2, h(x) = g(x). This means
that when a = 0, h(x) has four real roots, but by the time a = 2, h(x) is down
to only having two real roots. When center = 50 and width= 150 can you see any
changes to h(x) when you change the parameter a?
No, not really.
(e) Set center = −3.7 and width= 0.5. Comment on how changing the parameter
a will alter the graph of h(x). Determine at what approximate value of a the
function h(x) switches the number of real roots it has.
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It raise the function value of the minimum point until it is over the x-axis. The
approximate value is a = 1.
(f) Determine whether you agree or disagree with the following statement and explain
your choice. “Two polynomials with nearly identical coefficients will have nearly
identical roots.”
We disagree with this statement because a change of two bits of the non-highest
order terms removed two of the real roots. This small change caused a significant
difference in the roots of this system.
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