EXAMPLE 6 Discuss the curve y = 3x4 - 48x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If f(x) = 3x4 – 48x³, then f'(x) = 12x3 - 144x² = 12x?(x – 12) f"(x) = 36x2 – 288x = 36x(x – 8). 12 To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x = . To use the Second Derivative Test we evaluate f" at these critical numbers: f"(0) = Video Example ) f"(12) = and f"(12) > 0, f(12) = is a local minimum. Since f"(0) = , the Second Since f'(12) = Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x <0 and also for 0 < x < 12, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. [In fact, the expression for f'(x) shows that f decreases to the left of 12 and increase to the right of 12.] Since f"(x) = 0 when x = 0 or x = , we divide the real number line into intervals with these numbers as endpoints and complete the following chart. Interval f"(x) = 36x(x - 8) Concavity (-00, 0) upward (0.O) downward upward The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also, -12288) is an inflection point since the curve changes from concave downward to concave upward there. Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in the figure.
EXAMPLE 6 Discuss the curve y = 3x4 - 48x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. SOLUTION If f(x) = 3x4 – 48x³, then f'(x) = 12x3 - 144x² = 12x?(x – 12) f"(x) = 36x2 – 288x = 36x(x – 8). 12 To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x = . To use the Second Derivative Test we evaluate f" at these critical numbers: f"(0) = Video Example ) f"(12) = and f"(12) > 0, f(12) = is a local minimum. Since f"(0) = , the Second Since f'(12) = Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x <0 and also for 0 < x < 12, the First Derivative Test tells us that f does not have a local maximum or minimum at 0. [In fact, the expression for f'(x) shows that f decreases to the left of 12 and increase to the right of 12.] Since f"(x) = 0 when x = 0 or x = , we divide the real number line into intervals with these numbers as endpoints and complete the following chart. Interval f"(x) = 36x(x - 8) Concavity (-00, 0) upward (0.O) downward upward The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also, -12288) is an inflection point since the curve changes from concave downward to concave upward there. Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in the figure.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 15T
Related questions
Question
![EXAMPLE 6
Discuss the curve y = 3x4 – 48x³ with respect to concavity, points of inflection, and local maxima and
minima. Use this information to sketch the curve.
SOLUTION
If f(x) = 3x* – 48x³, then
-
12x3 – 144x2 = 12x2(x – 12)
36x(х — 8).
f'(x)
f"(x)
36x? – 288x
%3D
-
8
12
To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x =
To use the Second Derivative Test
we evaluate f" at these critical numbers:
f"(0)
%D
Video Example
f"(12)
Since f'(12) =
and f"(12) > 0, f(12) =
is a local minimum. Since f'"(0) =
the Second
Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for
0 < x < 12, the First Derivative Test tells us thatf does not have a local maximum or minimum at 0. [In fact, the
expression for f'(x) shows that f decreases to the left of 12 and increase to the right of 12.]
Since f"(x) = 0 when x = 0 or x =
we divide the real number line into intervals with these numbers as
endpoints and complete the following chart.
Interval
f"(x) = 36x(x –- 8) | Concavity
(-00, 0)
upward
+
downward
+
upward
The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also,
-12288)
is an inflection point since the curve changes from concave downward to concave upward there.
Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in the figure.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F982543f2-46cf-4f2e-9047-56f1e64038c1%2Fe18b6f18-fb1f-4fee-9dfe-f2c72bcf134e%2Fkcaurmb_processed.png&w=3840&q=75)
Transcribed Image Text:EXAMPLE 6
Discuss the curve y = 3x4 – 48x³ with respect to concavity, points of inflection, and local maxima and
minima. Use this information to sketch the curve.
SOLUTION
If f(x) = 3x* – 48x³, then
-
12x3 – 144x2 = 12x2(x – 12)
36x(х — 8).
f'(x)
f"(x)
36x? – 288x
%3D
-
8
12
To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x =
To use the Second Derivative Test
we evaluate f" at these critical numbers:
f"(0)
%D
Video Example
f"(12)
Since f'(12) =
and f"(12) > 0, f(12) =
is a local minimum. Since f'"(0) =
the Second
Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for
0 < x < 12, the First Derivative Test tells us thatf does not have a local maximum or minimum at 0. [In fact, the
expression for f'(x) shows that f decreases to the left of 12 and increase to the right of 12.]
Since f"(x) = 0 when x = 0 or x =
we divide the real number line into intervals with these numbers as
endpoints and complete the following chart.
Interval
f"(x) = 36x(x –- 8) | Concavity
(-00, 0)
upward
+
downward
+
upward
The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also,
-12288)
is an inflection point since the curve changes from concave downward to concave upward there.
Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in the figure.
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