Find the interval where the function Is concave up. Fin d the interval wnere the function is concave down. Step 1 -8x Since f'(x) then (x2 – 4)2' (x² – 4)²( )- (-8x) 2(x? – 4)| -8 f"(x) (x2 4) 80 3 x2 + 4 3 (x2 - 4) Step 2 The numerator is always positive, so the denominator determines the sign of f"(x). Thus, f"(x) is positive, V, if x2 > | , which means x > and so f(x) is concave up conclude that f"(x) is negative, and so f(x) is concave ---Select-- V or x < We can also 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.)
Find the interval where the function Is concave up. Fin d the interval wnere the function is concave down. Step 1 -8x Since f'(x) then (x2 – 4)2' (x² – 4)²( )- (-8x) 2(x? – 4)| -8 f"(x) (x2 4) 80 3 x2 + 4 3 (x2 - 4) Step 2 The numerator is always positive, so the denominator determines the sign of f"(x). Thus, f"(x) is positive, V, if x2 > | , which means x > and so f(x) is concave up conclude that f"(x) is negative, and so f(x) is concave ---Select-- V or x < We can also 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.)
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
Please help I need step 2
![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
\[
f''(x) = \frac{(x^2 - 4)(-8) - (-8x)\left(2(x^2 - 4)\left(2x\right)\right)}{(x^2 - 4)^3} = \frac{8\left(x^2 + 4\right)}{(x^2 - 4)^3}.
\]
**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
up
, if \( x^2 > \underline{\quad} \), which means \( x > \underline{\quad} \) or \( x < \underline{\quad} \). We can also conclude that \( f''(x) \) is negative, and so \( f(x) \) is concave
---
, if \( \underline{\quad} < x < \underline{\quad} \).
Therefore the interval where the function is concave up is as follows. (Enter your answer using interval notation.)
\[\underline{\quad\quad\quad}\]
The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
\[\underline{\quad\quad\quad}\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F892c0bd3-4224-4a72-a83b-d17d787f02ee%2Fcb87124f-c3d4-4610-81de-9787e1192eba%2F6hauxs6_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
\[
f''(x) = \frac{(x^2 - 4)(-8) - (-8x)\left(2(x^2 - 4)\left(2x\right)\right)}{(x^2 - 4)^3} = \frac{8\left(x^2 + 4\right)}{(x^2 - 4)^3}.
\]
**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
up
, if \( x^2 > \underline{\quad} \), which means \( x > \underline{\quad} \) or \( x < \underline{\quad} \). We can also conclude that \( f''(x) \) is negative, and so \( f(x) \) is concave
---
, if \( \underline{\quad} < x < \underline{\quad} \).
Therefore the interval where the function is concave up is as follows. (Enter your answer using interval notation.)
\[\underline{\quad\quad\quad}\]
The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
\[\underline{\quad\quad\quad}\]
Expert Solution
Step 1
Step by step
Solved in 2 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