Math 1000 Tutorial
Quiz 8
Week 8
The solution to Quiz 8 are below. I might be showing more steps than necessary, this is to aid your
understanding.
1. For f (x) = x3 − 12x, find the following and sketch the function.
Solution: Derivatives:
f 0 (x) = 3x2 − 12
f 00 (x) = 6x
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(a) Zeroes
Solution: (Zeroes are the x-intercepts)
f (x) = x3 − 12x
= x(x2 − 12) = 0
√
√
Thus, x = 0 or x2 − 12 = 0, meaning x = 0 or x = 12 or x = − 12. (The function value
at all of those points is zero, as the name indicates, and we solved for the points where this is
true!)
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(b) Critical points
Solution: Critical points are where the first derivative is zero or does not exist.
f 0 (x) = 3x2 − 12 = 3(x2 − 4) = 0
Thus x = 2 or x = −2. f (2) = 8 − 24 = −16 and f (−2) = −8 + 24 = 16.
There are no points at which the derivative doesn’t exist since it is a polynomial.
The critical points are (2, −16) and (−2, 16).
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(c) Inflection points
Solution: An inflection point can only occur if f 00 (x) = 0 or f 00 (x) does not exist (but similar
to the first derivative, this does not guarantee an inflection point, we also have to check if
concavity actually changes).
f 00 (x) = 6x = 0
Thus x = 0 and f (0) = 0. Since f 00 (x) is a straight line, it changes from negative to positive
here, thus concavity changes.
The inflection point therefore is (0, 0).
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(d) Areas of increase and decrease.
Solution: We have critical points at x = −2 and x = 2. Thus we need to check if the first
derivative is positive or negative on the intervals (−∞, −2), (−2, 2) and (2, ∞). We can choose
any x-value in those intervals:
f 0 (−3) = 27 − 12 > 0
f 0 (0) = −12 < 0
f 0 (3) = 27 − 12 > 0
Thus, the function is increasing on the intervals (−∞, −2) and (2, ∞) and decreasing on the
interval (−2, 2).
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(e) Areas of concave up and concave down
Solution: The second derivative is zero at x = 0, so we need to check concavity on the intervals
(−∞, 0) and (0, ∞).
f 00 (−1) = −6 < 0
f 00 (1) = 6 > 0
Thus the function is concave up on the interval (0, ∞) and concave down on the interval (−∞, 0).
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(f) Vertical and horizontal asymptotes
Solution: There are no vertical or horizontal asymptotes
1 (bonus)
i. Explain why.
Solution: There are no vertical asymptotes since f (x) is defined everywhere. There are
no horizontal asymptotes since
lim f (x) = lim x3 − 12x = lim x(x2 − 12) = ∞ · ∞ = ∞
x→∞
x→∞
x→∞
lim f (x) = lim x3 − 12x = lim x(x2 − 12) = −∞ · ∞ = −∞
x→∞
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x→∞
x→∞
(g) Local max and min
Solution: There is a local maximum at x = −2 since the function is concave down at this
point (or since the function changes from increasing to decreasing at this point).
There is a local minimum at x = 2 since the function is concave up at this point (or since the
function changes from decreasing to increasing at this point).
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(h) Sketch.