yEXAMPLE 6Discuss the curve y = 3x* – 48x with respect to concavity, points of inflection, and local maxima and minima. Use this information tosketch the curve.SOLUTIONIf f(x) = 3x4 - 48x³, then144x2 = 12x2(x – 12)288х 3D 36x(х — 8).f'(x)12x3f"(x) = 36x2812To 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 criticalnumbers:f"(0)f"(12)Video ExampleSince f'(12)and f"(12) > 0, f(12) =is a local minimum. Since f"(0)the Second Derivative Test gives no information%3Dabout 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 localmaximum or minimum at 0. [In fact, the expression for f'(x) shows that f decreases to the left of 12 and increase to theof 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 followingchart.Intervalf"(x)36x(х — 8) Concavity(-0, 0)upwarddownward+upwardThe point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also,-12288is aninflection 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.

If fx=3x4-48x3, thenf'x=12x3-144x2=12x2x-12f''x=36x2-288x=36xx-8 To find the critical numbers we set…

EXAMPLE 6 Discuss the curve y = 3x* - 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* – 48x², then f'(x) = 12x3 - 144x² = 12x²(x – 12) f"(x) = 36x2 – 288x = 36x(x – 8). To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x = | To use the Second Derivative Test we evaluate " at these critical numbers: f"(0 : f"(12) Since f'(12) = and f"(12) > 0, (12) = O is a local minimum. Since f"(0) =O, the Second Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for 0

EXAMPLE 6 Discuss the curve y = 3x* - 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* – 48x², then f'(x) = 12x3 - 144x² = 12x²(x – 12) f"(x) = 36x2 – 288x = 36x(x – 8). To find the critical numbers we set f'(x) = 0 and obtain x = 0 and x = | To use the Second Derivative Test we evaluate " at these critical numbers: f"(0 : f"(12) Since f'(12) = and f"(12) > 0, (12) = O is a local minimum. Since f"(0) =O, the Second Derivative Test gives no information about the critical number 0. But since f'(x) < 0 for x < 0 and also for 0

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Concept explainers

Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

Question

y
EXAMPLE 6
Discuss the curve y = 3x* – 48x 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
144x2 = 12x2(x – 12)
288х 3D 36x(х — 8).
f'(x)
12x3
f"(x) = 36x2
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)
f"(12)
Video Example
Since f'(12)
and f"(12) > 0, f(12) =
is a local minimum. Since f"(0)
the Second Derivative Test gives no information
%3D
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
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(х — 8) Concavity
(-0, 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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Discuss the curve y = 3x4 - 12x³ with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve. Solution If f(x)= 3x4 12x3, then f'(x) = 12x³ - 36x² = 12x²(x - 3) f"(x) = 36x²72x = 36x(x - 2). 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 the following critical numbers. f"(0) = f"(3) = Since f'(3) = and f"(3) > 0, f(3) = is a local minimum. Since f"(0) = expression for f'(x) shows that f decreases to the left of 3 and increases to the right of 3.] Since f"(x) = 0 when x = 0 or x = Interval (-00, 0) ,00) F"(x) = 36x(x - 2) Concavity + upward we divide the real line into intervals with these numbers as endpoints and complete the following chart. downward -48 -81 upward The point (0, 0) is an inflection point since the curve changes from concave upward to concave downward there. Also, Using the local minimum, the intervals of concavity, and the inflection…
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