11/3/2016
Activity 4: Solutions
MATH 120, Fall 2016
1. Suppose f (x) is a function with f 0 (x) = (x − 1)(x + 1)(x − 7) = x3 − 7x2 − x + 7.
(a) Find the inflection points of f (x).
With f 0 (x) = x3 −7x2 −x+7, we note that f 00 (x) = 3x2 −14x−1. We need to identify possible candidates
for sign change of f 00 (these are called hypercritical points); if f 00 (x) changes sign at x = c (and x = c is
in the domain of f (x)) then f (x) changes in concavity at x = c.
Using the quadratic formula, we note that the zeros of f 00 (x) are
√
7 ± 2 13
x=
≈ −.070, 4.737.
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We do not construct a number line here, but if we did, it would read “Sign of f 00 ”. We are checking the
signs at points on either side of the hypercritical values. Let us consider x = −1, 0, 5.
f 00 (−1) = 16 > 0
f 00 (0) = − 1 < 0
f 00 (5) = 4
Thus the sign of f 00 (x) changes at each hypercritical value, and, as discussed above, this corresponds to
a change of concavity. Thus each hypercritical value is the location of an inflection point. The inflection
points of f (x) are therefore:
√ !!
√
√ !!
√
7 + 2 13
7 − 2 13
7 + 2 13
7 + 2 13
,f
,
,f
.
3
3
3
3
(b) Use the first derivative test to find the places where local extrema occur for f (x). Classify the extrema
(don’t give the y-values).
The first derivative is given to us in factored form. We can immediately see that there are no x-values for
which f 0 (x) DNE. However, there are three zeros of f 0 (x): x = −1, 1, 7. These are the critical values. To
use the FDT, we are looking for sign changes of f 0 (x) at these CVs. We need, in order to use the FDT,
to test values on either side of these CVs. Let us choose x = −2, 0, 2, 10 and focused on the factored form
of f 0 (x) above.
f 0 (−2) = (−)(−)(−) < 0
f 0 (0) = (−)(+)(−) > 0
f 0 (2) = (+)(+)(−) < 0
f 0 (10) =(+)(+)(+) > 0
Thus we see sign changes at each of the critical values. Hence, by the FDT we have:
• Since f 0 (x) goes from − → + at x = −1 (f (x) goes from decreasing to increasing) we know that
(−1, f (−1)) is a local minimum of f (x).
• Since f 0 (x) goes from + → − at x = −1 (f (x) goes from increasing to decreasing) we know that
(1, f (1)) is a local maximum of f (x).
• Since f 0 (x) goes from − → + at x = −7 (f (x) goes from decreasing to increasing) we know that
(7, f (7)) is a local minimum of f (x).
(c) Use the second derivative test to find the places where local extrema occur for f (x). These should agree
with the previous part.
We test each of the CVs (x = −1, 1, 7) in the second derivative. The sign will determine if they are
extrema (and what type).
f 00 (−1) = 16 > 0
f 00 (1) = − 12 < 0
f 00 (7) = BIG > 0
Thus, by the SDT we have:
• Since x = −1 is a CV and f 00 (−1) > 0, we know that (−1, f (−1)) is a local minimum of f (x).
• Since x = 1 is a CV and f 00 (1) < 0, we know that (1, f (1)) is a local maximum of f (x).
• Since x = 7 is a CV and f 00 (7) > 0, we know that (7, f (7)) is a local minimum of f (x).
(d) Where do the local extrema occur for f 0 (x)? Explain.
Since f, f 0 , and f 00 are all continuous, we know that the inflection points of f (x) are the same as the local
extrema of f 0 (x)...why? Well, if x = c is an inflection point of f (x) then the concavity of f changes at
x = c. This means f 00 (x) changes sign at x = c. This is equivalent to [f 0 (x)]0 changes sign. Thus applying
the FDT to the function f 0 (x), we conclude that if f 00 (x) changes sign at x = c, then [f 0 (x)]0 does as well.
√
7 ± 2 13
0
.
Hence, the local extrema for f (x) occur at x =
3
2. Consider the graph below.
2
Approximating to the best of your ability, give the x-values where:
(a) f 0 (x) > 0
This occurs whenever f is increasing. This seems to be: (−4.5, −2) ∪ (3 − , 3 + ) ∪ (7.25, ∞)
(b) f is concave down
(−∞, −6) ∪ (−3, 0) ∪ (0, 1 + ) ∪ (3, 5 + )
(c) f 0 (x) = 0 or DNE
x = −4.5, −2, 0, , 3 − , 3 + , 7 + (d) inflection points of f
x = −6, −3, 1 + , 7.25
(e) local extrema for f occur
x = −4 − , −2, 0, , 3 − , 3 + , 7 + (f) absolute extrema on [−8, 8] occur
x = 3 − is where an absolute minimum occurs; x = 3 + is where absolute maximum occurs.
(g) absolute extrema on [−1, 5] occur
On this interval, the absolute extrema occur in the same places as before.
(h) local extrema for f 0 occur
3. Given the following information about f (x), sketch the function described below.
(Skeleton)
• The domain of f (x) is (−∞, −2) ∪ (−2, 1) ∪ (1, ∞).
• f (−3) = f (−1) = f (2) = f (4) = 0.
• f (0) = −1, f (−5) = −3, f (−4) = −4, f (3) = −1.
•
lim f (x) = −2,
x→−∞
• lim f (x) = +∞,
x→−2
• For x >>> 0,
lim f (x) = +∞.
x→∞
lim f (x) = −∞,
x→1−
f (x) ≈
lim f (x) = +∞.
x→1+
6
x − 6.
5
(First Deriv.)
• f 0 (x) > 0 on (−4, −2) ∪ (3, ∞); f 0 (x) < 0 on (−∞, −4) ∪ (−2, 1) ∪ (1, 3).
• f 0 (x) = 0 for x = −4, 0, 3; f 0 (x) DN E for x = −2, 1.
(Second Deriv.)
• f 00 (x) > 0 on (−5, −2) ∪ (−2, 0) ∪ (1, ∞); f 00 (x) < 0 on (−∞, −5) ∪ (0, 1).
• f 00 (x) = 0 for x = −5, 0; f 00 (x) DN E for x = −2, 1.
There are many correct answers to this question.
4. Sketch the following curves by finding: the maximal domain of the function, vertical and horizontal asymptotes,
intercepts (where possible), intervals of increase/decrease, intervals of concave up/down, local extrema, and
inflection points. Label all relevant points.
(a) y = −x3 + 4x2 − x − 6
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(b) f (x) =
x
x2 − 1
(c) g(x) = ln(1 − ln(x))
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