Download Curve sketching - Douglas Weathers

MATH 132
CURVE SKETCHING
1. Sketch f (x) = x2 − 5x − 14 according to the following criteria.
Domain: All real numbers
Asymptotes: None, since f is a polynomial
End behavior: For large values of |x|, the graph of f looks like x2 (highest
term)
Intercepts: The y-intercept is f (0) = −14. The x-intercepts are
x2 − 5x − 14 = 0
(x − 7)(x + 2) = 0
x = 7, −2
First derivative and critical numbers: The first derivative is f 0 (x) = 2x − 5.
Solve f 0 (x) = 0:
2x − 5 = 0 ⇒ x = 5/2
Since f 0 is a polynomial, it is never undefined. The only critical number is
c = 5/2.
Second derivative and inflection numbers: The second derivative is f 00 (x) =
2, which is never zero. There are therefore no inflection points. Since f 00 = 2 >
0, the graph of f is always concave upwards, so the second derivative test tells
us that f (5/2) is a local minimum.
Sign chart:
00
f
f0
f
(−∞, −2)
+
+
(−2, 25 )
+
-
( 52 , 7)
+
+
-
40
20
−5
5
−20
(7, ∞)
+
+
+
2. Sketch g(x) = x/(x2 − 9) according to the following criteria.
Domain: All x such that
x2 − 9 6= 0
(x − 3)(x + 3) 6= 0
x 6= −3, 3
Asymptotes & end behavior: Vertical asymptotes at x = −3, 3. The end
behavior ( lim ) of g is x/x2 = 0, so g has a horizontal asymptote at y = 0.
x→∞
Intercepts: The y-intercept is g(0) = 0/(−9) = 0. The x-intercept is found
when the numerator is zero, so x = 0. Therefore the only intercept is (0, 0).
First derivative and critical numbers: According to the quotient rule,
−(x2 + 9)
x2 − 9 − 2x2
=
2
2
(x − 9)
(x2 − 9)2
g 0 (x) =
The derivative is never zero since x2 +9 = 0 has no real solutions. It is undefined
when x = −3, 3, but those numbers are not in our domain. Hence there are no
critical numbers.
Second derivative and inflection numbers:
g 00 (x) = −
2x(x2 − 9)2 − 2(2x)(x2 + 9)
(x2 − 9)4
Since the derivative is always defined on our domain, inflection points will
appear where g 00 (x) = 0. Solve:
(x2 − 9) 4x − 2x(x2 − 9) = 0
Since x2 − 9 6= 0 on our domain, solve
4x − 2x(x2 − 9) = 0
− 2x3 − 18x = 0
x(x2 + 9) = 0
Since x2 + 9 = 0 has no real solutions, the inflection number is x = 0.
Sign chart:
g 00
g0
g
(−∞, −3)
-
(−3, 0)
+
+
(0, 3)
-
(3, ∞)
+
+
10
5
−4
−2
2
−5
−10
4