Download CURVE SKETCHING I. THE FIRST DERIVATIVE TEST: a

CURVE SKETCHING
I. THE FIRST DERIVATIVE TEST:
Let's take an arbitrary function like the one whose graph is given below:
As x goes from a to p, the graph rises as x moves to the right towards the interval P, (a, p) and
the value of f increases. We say that the function is increasing on the open interval (a, p). As x
continues to move towards q the graph falls and the value of f decreases, we would say that the
open interval (p,q) is decreasing. Since f(r) ≥ f(x) for all x in (a, b), we say that the function f has
its maximum value on (a, b) when x = r and that f(r) is the maximum value of f on interval
(a, b). Notice also that f(q) ≤ f(x) for all x in interval (a, b), we say that the function ƒ has its
minimum value on interval (a, b) when x = q and that f(q) is the minimum value of f on
interval (a, b).
Example 1: Find the critical numbers of f if
f(x) = 2x3 - 6x2 - 18x + 7
f '(x)= 6x2 - 12x -18
6x2 - 12x -18 = 0
(6x + 6) (x - 3) = 0
6x + 6 = 0
or
x = -1
or
x-3=0
x=3
find f '(x) and set it = 0
factor
solve for x
So, the critical numbers are x = -1 and x = 3.
Example 2: Find the intervals on which ƒ is increasing if
f(x) = 2x3 - 6x2 - 18x + 7
f is increasing if f ' is positive, so we need to find when f '(x) >0 or 6x2 - 12x - 18 > 0. From Example 1 we
know that the critical points are x = -1 and x = 3.
We determine the sign of f ' on the intervals between critical points by using test numbers that fall to the
left, between, and to the right of our critical points and plug them into f '(x).
(a) From the interval (- ∞ , -1) we choose x = -2 as a test number:
f ' (-2) = 6(-2)2 - 12(-2) - 18 = 30 > 0
f is increasing on (- ∞ , -1)
(b) From the interval (-1, 3) we choose x = 0 as a test number:
f ' (0) = 6(0)2 - 12(0) - 18 = -18 < 0
f is decreasing on (-1, 3)
(c) From the interval (3, ∞ ) we choose x = 4 as a test number:
f in increasing on (3, ∞ )
f ' (4) = 6(4)2 -12(4) - 18 = 30 > 0
From (a), (b), and (c) we see that f is increasing on the intervals (- ∞ , -1) and (3, ∞ ).
Example 3: Find the relative extreme values of f (maximum and minimum) if
f(x) = 2x3 - 6x2 - 18x + 7
Make the following chart from the results of Example 2:
+
Sign of f'
-1
+
3
Notice that f ' changes sign from (+) to (-) at x = -1 ⇒ f has a relative maximum at x = -1
The value of the relative maximum is f (-1) = 17
f ' changes sign from (-) to (+) at x = 3 ⇒ f has a relative minimum at x = 3
The value of the relative minimum is f (3) = -47
Example 4: Sketch the graph of f if
f (x) = 2x3 - 6x2 - 18x + 7
•
•
•
•
•
ALWAYS start left to right when graphing.
Plot the maximum and minimum points: (-1, 17) and (3, -47)
f is increasing on (- ∞ , -1); the graph increases until x = -1
f is decreasing on (-1, 3); the graph decreases until x = 3
f is increasing again on (3, ∞ ); the graph increases until x = ∞ :
II. CONCAVITY AND THE SECOND DERIVATIVE TEST:
A graph can be either increasing or decreasing, it could concave up, down, both or neither:
The concavity of the curve has nothing to do with whether the function is increasing or
decreasing. See above.
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Inflection Point:
A point is said to be an inflection point if the graph changes concavity at that point.
The Second Derivative Test:
If f '(c) = 0 (c is a critical number) and
f ''(c) > 0, then f has a relative minimum at x = c and is concave up
f ''(c) < 0, then f has a relative maximum at x = c and is concave down
f ''(c) = 0, the c is an inflection point and changes concavity at x = c.
Example 5: Use the second derivative test to find the relative extreme values of f if
f (x) = 2x3 - 6x2 - 18x + 7
f '(x) = 6x2 - 12x - 18
The critical numbers are x = -1 and x = 3
f ''(x) = 12x -12
(see Example 1)
Now evaluate f ''(x) at the critical numbers:
f ''(-1) = 12(-1) -12 = -24 < 0 ⇒ relative maximum at x = -1
f ''(3) = 12(3) -12 = 24 > 0 ⇒ relative minimum at x = 3
(see Example 2)
Example 6: Use the second derivative test to find the relative extreme values and the inflection
points of f if
f (x) = 2 - 9x + 6x2 - x3
f '(x)= -9 + 12x - 3x2
-9 + 12x - 3x2 = 0
-3(x-3)(x-1) = 0
x=3
or
x=1
find f '(x) and set it = 0
factor
solve for x
The critical numbers are: x = 1 and x = 3.
Sign-of-f'
-
+
-
1
3
f is decreasing on both (- ∞ , 1) and (3, ∞ ) and increasing on (1, 3).
f (1) = -2 and f(3) = 2
Plot the points (1, 2) and (3, 2) on the graph.
To discuss concavity look at f '':
f ''(x) = 12 - 6x = 0
x=2
• We determine the sign of f '' by choosing points that lie to the left, between or to the right of our
inflection points and plugging them into f ''(x).
• For interval (- ∞, 2) we choose x = 0; f ''(0) = 12-6(0) = 12 > 0.
•
•
•
•
•
For interval ( 2, ∞ ) we choose x = 4; f ''(4) = 12-6(4) = -12< 0.
Sign-of-f ''
+
2
•
•
•
the graph is concave up on (- ∞ , 2) and concave down on (2, ∞ )
x = 2 is an inflection point; f(2) = 2 - 9(2) + 6(2)2 - (2)3 =0:
x = 2 is also the x-intercept; f(2) = 0
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III. ASYMPTOTES:
Asymptotes exist only when the function consists of fractional polynomials. We have three types
of asymptotes; vertical, horizontal and oblique (slant):
Vertical Asymptotes:
Whatever number makes the denominator zero is a vertical asymptote:
x 3 − 5x 2 + 6 x + 5
i.e., f(x) =
, factor x2 – 3x -4 down to (x – 4)(x + 1),
2
x − 3x − 4
x = 4 and x = -1 are vertical asymptotes.
Horizontal Asymptotes
Look at the highest power of the numerator (call it t for top) and the highest
power of the denominator (call it b for bottom); three cases are considered.
Case 1: b > t
The line y = 0 is a horizontal asymptote.
Case 2.: b = t
Divided the coefficient of the xt term by the coefficient of the xb term; the result is the
horizontal asymptote.
Case 3: b + 1 = t
Divide the numerator by the denominator and the quotient is called an oblique (slant)
asymptote.
Otherwise:
No asymptotes exist.
Example 7: Find the relative extreme values, the inflection points of f and sketch the graph of:
f (x) =
6− x
x2
f '(x) = x − 12 = 0
x3
x - 12 = 0
x = 12
•
find f '(x) and set it = 0
set the numerator only = 0
solve for x
The critical numbers are: x = 12 and x = 0
(denominator cannot = 0, so that value is
considered a critical number)
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• f is increasing on both (- ∞ , 0) and (12, ∞ ) and decreasing on (0, 12).
• f(0) = undefined and f(12) = -6/144
• the point (12, -6/144) is a relative minimum.
• Plot the point (12, -6/144) on the graph.
• To discuss concavity look at f '':
f ''(x)= − 2 x + 36 = 0
x4
x = 18 and again x = 0 are critical numbers
+
Sign-of-f ''
•
•
•
•
•
+
0
18
the graph is concave up on (- ∞ , 0) and (0, 18) and concave down on (18, ∞ )
x = 18 is the only inflection point:
Vertical asymptotes: x = 0
(denominator = 0)
Horizontal asymptotes: y = 0
(second case)
and finally the graph:
Revised: Spring 2005
Created in 1994 by Ziad Diab
STUDENT LEARNING ASSISTANCE CENTER (SLAC)
Texas State University-San Marcos
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