EXAMPLE 2 Find the local minimum and maximum values of the function below. f(x) 3x4 – 8x3 – 90x² + 6 Video Example SOLUTION f'(x) = 12x3 – 24x² – 180x = 12x(x – 5)(x + 3) From the chart Interval 12x x - 5 x + 3 f'(x) x < -3 decreasing on (-∞, –3) -3 < x < 0 + + increasing on (-3, 0) 0 < x < 5 + decreasing on (0, 5) x > 5 + + + increasing on (5, ∞) we see that f '(x) changes from negative to positive at -3, so f(-3) = is a local minimum value by the First Derivative Test. Similarly, f'(x) changes from negative to positive at 5, so f(5) : is a local minimum value. And, f(0) = is a local maximum value because f'(x) changes from positive to negative at 0. +

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter1: Expressions And Functions
Section1.8: Interpreting Graphs Of Functions
Problem 19PPS
icon
Related questions
icon
Concept explainers
Question
EXAMPLE 2
Find the local minimum and maximum values of the function below.
f(x) = 3x4 – 8x³ – 90x² + 6
-
Video Example
SOLUTION
f'(x) = 12x3 – 24x2 – 180x =
12x(x – 5)(x + 3)
From the chart
Interval
12x
X - 5
х+ 3
f'(x)
f
X < -3
decreasing on (-∞, -3)
-3 < x < 0
+
increasing on (-3, 0)
0 < x < 5
+
+
decreasing on (0, 5)
x > 5
+
+
+
+
increasing on (5, ∞)
we see that f '(x) changes from negative to positive at -3, so f(-3) =
is a local minimum value by the
First Derivative Test.
Similarly, f'(x) changes from negative to positive at 5, so f(5) =
is a local minimum value.
And, f(0)
is a local maximum value because f'(x) changes from positive to negative at 0.
Transcribed Image Text:EXAMPLE 2 Find the local minimum and maximum values of the function below. f(x) = 3x4 – 8x³ – 90x² + 6 - Video Example SOLUTION f'(x) = 12x3 – 24x2 – 180x = 12x(x – 5)(x + 3) From the chart Interval 12x X - 5 х+ 3 f'(x) f X < -3 decreasing on (-∞, -3) -3 < x < 0 + increasing on (-3, 0) 0 < x < 5 + + decreasing on (0, 5) x > 5 + + + + increasing on (5, ∞) we see that f '(x) changes from negative to positive at -3, so f(-3) = is a local minimum value by the First Derivative Test. Similarly, f'(x) changes from negative to positive at 5, so f(5) = is a local minimum value. And, f(0) is a local maximum value because f'(x) changes from positive to negative at 0.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Knowledge Booster
Application of Differentiation
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Glencoe Algebra 1, Student Edition, 9780079039897…
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage