Find the interval where the function is concave up. Find the interval where the function is concave down. Step 1 -8x Since f'(x) then %3D (x² – 4)2'
Find the interval where the function is concave up. Find the interval where the function is concave down. Step 1 -8x Since f'(x) then %3D (x² – 4)2'
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...
Related questions
Question
i need step 2 please
![## Find the interval where the function is concave up. Find the interval where the function is concave down.
### Step 1
Since \( f'(x) = \frac{-8x}{(x^2 - 4)^2} \), then:
\[
f''(x) = \frac{(x^2 - 4)^2 (-8) - (-8x) \left[ 2(x^2 - 4)(2x) \right]}{(x^2 - 4)^4}
\]
Simplifying this expression:
- The numerator expands and simplifies to:
\[
8 \cdot \left( 3x^2 + 4 \right)
\]
- The denominator is:
\[
(x^2 - 4)^3
\]
### Step 2
The numerator is always positive, so the denominator determines the sign of \( f''(x) \). Thus, \( f''(x) \) is positive, and so \( f(x) \) is concave \(\_\_\_\), if \( x^2 > \_\_\_\), which means \( x > \_\_\_\) or \( x < \_\_\_\). We can also conclude that \( f''(x) \) is negative, and so \( f(x) \) is concave \(\_\_\_\), if \(\_\_\_\) < \( x \) < \(\_\_\_\).
Therefore, the interval where the function is concave up is as follows. (Enter your answer using interval notation.)
\[ \_\_\_ \]
The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
\[ \_\_\_ \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892c0bd3-4224-4a72-a83b-d17d787f02ee%2F524f491f-f6a1-4489-bc8b-ab7096e8d35d%2Fv3pylk6_processed.png&w=3840&q=75)
Transcribed Image Text:## Find the interval where the function is concave up. Find the interval where the function is concave down.
### Step 1
Since \( f'(x) = \frac{-8x}{(x^2 - 4)^2} \), then:
\[
f''(x) = \frac{(x^2 - 4)^2 (-8) - (-8x) \left[ 2(x^2 - 4)(2x) \right]}{(x^2 - 4)^4}
\]
Simplifying this expression:
- The numerator expands and simplifies to:
\[
8 \cdot \left( 3x^2 + 4 \right)
\]
- The denominator is:
\[
(x^2 - 4)^3
\]
### Step 2
The numerator is always positive, so the denominator determines the sign of \( f''(x) \). Thus, \( f''(x) \) is positive, and so \( f(x) \) is concave \(\_\_\_\), if \( x^2 > \_\_\_\), which means \( x > \_\_\_\) or \( x < \_\_\_\). We can also conclude that \( f''(x) \) is negative, and so \( f(x) \) is concave \(\_\_\_\), if \(\_\_\_\) < \( x \) < \(\_\_\_\).
Therefore, the interval where the function is concave up is as follows. (Enter your answer using interval notation.)
\[ \_\_\_ \]
The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
\[ \_\_\_ \]
Expert Solution
Step 1
for concave up f"(x) >0
and for concave down f"(x) <0
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning