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Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
1. The First Derivative
Consider the function f (x) = x3 − 3x2 − 9x + 10
The first derivative of f (x) is
f 0 (x) = 3x2 − 6x − 9 = 3(x2 − 2x − 3) = 3(x + 1)(x − 3)
The two critical (flat) points for f (x) are found by solving
f 0 (x) = 3(x + 1)(x − 3) = 0
Thus the two critical points are x = −1 and x = 3.
Using the critical points you can create a sign table for f 0 (x)
f 0 (x)
−1
(−)(−) = (+)
3
(+)(−) = (−)
Thus we see that f (x) is increasing on the interval (−∞, −1).
f (x) is decreasing on the interval (−1, 3).
f (x) is increasing on the interval (3, ∞).
From this we can conclude that:
The critical point (−1, f (−1)) = (−1, 15) is a local maximum.
The critical point (3, f (3)) = (3, −17) is a local minimum.
The graph of f (x) is
(+)(+) = (+)
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Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
2. The Second Derivative
Consider the function f (x) = x4 − 4x3 − 48x2 + 15x − 23
The first derivative of f (x) is
f 0 (x) = 4x3 − 12x2 − 96x + 15
The second derivative of f (x) is
f 00 (x) = 12x2 − 24x − 96 = 12(x2 − 2x − 8) = 12(x + 2)(x − 4)
The two partition numbers (possible inflection points) of f 00 (x) are found by solving
f 00 (x) = 12(x + 2)(x − 4) = 0
Thus x = −2 or x = 4.
The sign table for f 00 (x) (by using the above partition numbers) is
f 00 (x)
−2
(−)(−) = (+)
4
(+)(−) = (+)
(+)(+) = (+)
Thus f (x) is concave up on the interval (−∞, −2).
f (x) is concave down on the interval (−2, 4).
f (x) is concave up on the interval of (4, ∞).
Since concavity changes at each of the partition numbers it follows that:
The point (−2, f (−2)) = (−2, −197) and (4, f (4)) = (4, −731) are inflection points.
The graph of f (x) is
2
Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
3
3. Polynomial Functions
Consider the polynomial function P (x) = −4x(x + 2)3 = −4x4 − 24x3 − 48x2 − 32x
(a) The x-intercepts (zeros) of P (x) are found by solving
P (x) = −4x(x + 2)3 = 0
Thus the two zeros are x = −2 and x = 0, so the two x-intercepts are
(−2, 0)
and
(0, 0)
The sign table (use the zeros as partition numbers) for P (x) is
P (x)
−2
(−)(−)(−)3 = (−)
0
(−)(+)(−)3 = (+)
(−)(+)(+)3 = (−)
Thus P (x) is below the axis on the interval (−∞, −2).
P (x) is above the axis on the interval (−2, 0).
P (x) is below the axis on the interval (0, ∞).
The y-intercept of P (x) is (0, P (0)) = (0, 0)
(b) The first derivative of P (x) is
1
P 0 (x) = [−4](x + 2)3 + [3(x + 2)2 (1)](−4x) = −4(x + 2)2 [(x + 2) + 3x] = −4(x + 2)2 (4x + 2) = −16(x + 2)2 (x + )
2
The critical points for P (x) are found by solving
1
P 0 (x) = −16(x + 2)2 (x + ) = 0
2
Thus the two critical points are x = −2 and x = −1/2.
The sign table for P 0 (x) is
0
P (x)
−2
(−)(−) (−) = (+)
2
2
−1
2
(−)(+) (−) = (+)
(−)(+)2 (+) = (−)
Thus P (x) is increasing on the interval (−∞, −2).
P (x) is increasing on the interval (−2, −1/2).
P (x) is decreasing on the interval (−1/2, ∞).
Thus we know that the critical point (−2, f (−2)) = (−2, 0) is a flat point that is not a max or min.
The point (−1/2, f (−1/2)) = (−1/2, 27/4) is a local maximum.
Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
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(c) The second derivative of P (x) is
1
1
P 00 (x) = [−32(x + 2)(1)](x + ) + [1](−16(x + 2)2 ) = −16(x + 2)[2(x + ) + (x + 2)]
2
2
= −16(x + 2)[2x + 1 + x + 2] = −16(x + 2)(3x + 3) = −48(x + 2)(x + 1)
The possible inflection points (partition numbers for P 00 (x)) are found by solving
P 00 (x) = −48(x + 2)(x + 1) = 0
Thus x = −2 or x = −1.
The sign table for P 00 (x) is
P 00 (x)
−2
(−)(−)(−) = (−)
−1
(−)(+)(−) = (+)
(−)(+)(+) = (−)
Thus P (x) is concave down on the interval (−∞, −2).
P (x) is concave up on the interval (−2, −1).
P (x) is concave down in the interval (−1, ∞).
The two points (−2, f (−2)) = (−2, 0) and (−1, f (−1)) = (−1, 4) are inflection points since the concavity changes.
(d) The graph of P (x) is
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Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
4. Rational Functions
Consider the rational function R(x) =
2x2 − 6x
2x(x − 3)
=
x2 + x − 6
(x + 3)(x − 2)
(a) The domain of R(x) is all x such that
x2 + x − 6 = (x + 3)(x − 2) 6= 0
Thus x 6= −3 and x 6= 2. So the domain is
(−∞, −3) ∪ (−3, 2) ∪ (2, ∞)
The zeros of R(x) occur when R(x) = 0 or
2x2 − 6x = 2x(x − 3) = 0
Thus x = 0 or x = 3. Thus we have two x-intercepts (0, 0) and (3, 0).
The y-intercept is (0, R(0)) = (0, 0).
The sign table for R(x) (using the zeros and domain) is
R(x)
−3
(−)(−)
= (+)
(−)(−)
0
(−)(−)
= (−)
(+)(−)
2
(+)(−)
= (+)
(+)(−)
3
(+)(−)
= (−)
(+)(+)
(+)(+)
= (+)
(+)(+)
Thus we have that R(x) is above the axis on the interval (−∞, −3).
x = −3 is a vertical asymptote with lim − R(x) = +∞ and lim + R(x) = −∞.
x→−3
x→−3
R(x) is below the axis on the interval (−3, 0).
(0, 0) is an x-intercept.
R(x) is above the axis on the interval (0, 2).
x = 2 is a vertical asymptote with lim− R(x) = +∞ and lim+ R(x) = −∞.
x→2
x→2
R(x) is below the axis on the interval (2, 3).
(3, 0) is an x-intercept.
R(x) is above the axis on the interval (3, ∞).
(b) The first derivative is
[4x − 6](x2 + x − 6) − [2x + 1](2x2 − 6x)
(x2 + x − 6)2
3
2
(4x + 4x − 24x − 6x2 − 6x + 36) − (4x3 − 12x2 + 2x2 − 6x)
=
(x2 + x − 6)2
8x2 − 24x + 36
=
(x2 + x − 6)2
R0 (x) =
The critical points of R(x) occur when R0 (x) = 0 or
8x2 − 24x + 36 = 0
The discriminant of the quadratic formula is
b2 − 4ac = (−24)2 − 4(8)(36) = −576 < 0
So there are no real zeros and no critical points. Thus the numerator is positive for all real values of x (since it is
a parabola that opens upward with no zeros or intercepts).
Math 160 - Assignment 5 Solutions - Summer 2012 - BSU - Jaimos F Skriletz
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Putting this together we see that the numerator of R0 (x) is always positive and the denominator of R0 (x) is a
square, so it to is always positive. Thus R0 (x) > 0 for all x in the domain. This gives the following sign table for
R0 (x).
R0 (x)
−3
(+)
Thus R(x) is increasing on the interval (−∞, −3).
R(x) is increasing on the interval (−3, 2).
R(x) is increasing on the interval (2, ∞).
(c) The graph of R(x) is
2
(+)
(+)