Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
A review exercise set focused on finding open intervals where functions are increasing or decreasing, as well as locating and evaluating relative extrema. The exercises involve finding critical numbers and testing intervals to determine the sign of the derivative.
Typology: Study notes
1 / 9
This page cannot be seen from the preview
Don't miss anything!
Exercise 1: Find the open intervals on the graph where the given function is increasing or decreasing.
Increasing:
Decreasing:
Exercise 2: Find the open intervals where the given function is increasing or decreasing.
y = x 3 + x 2 - 8x + 5
Exercise 3: Find the locations and values of all relative extrema for the function in the graph below.
f(x) = (x + 1) / (x 2 + 2x + 2)
Exercise 4: Find the values of any relative extrema for the given function.
y = x 2/3^ (x 2 - 4)
Exercise 1: Find the open intervals on the graph where the given function is increasing or decreasing.
Exercise 2: Find the open intervals where the given function is increasing or decreasing.
y = x 3 + x 2 - 8x + 5
Find the first derivative
y = x 3 + x 2 - 8x + 5 y' = 3x 2 + 2x - 8
Find the critical numbers
0 = 3x 2 + 2x - 8 0 = (3x - 4)(x + 2) 3x - 4 = 0 or x + 2 = 0
x =
or x = -
Exercise 2 (Continued):
Test each interval to determine the sign of the derivative
Test points: -4, 0, 2
Decreasing:
Exercise 3: Find the locations and values of all relative extrema for the function in the graph below.
f(x) = (x + 1) / (x 2 + 2x + 2)
Exercise 4 (Continued):
Find the critical numbers
x 5/3^ -
x -1/
x -1/3(x 2 - 1)
0 = x -1/3(x 2 - 1)
Set each factor equal to zero and solve for x
x -1/3^ = 0
3
The fraction cannot be zero because the numerator is a constant but we can still get a critical number by setting the denominator equal to zero.
x = 0
x 2 - 1 = 0 x 2 = 1 x = ± 1
The critical numbers are -1, 0, and 1
Evaluate the function at the critical numbers
y = x 2/3^ (x 2 - 4); when x = - y = (-1) 2/3^ ((-1) 2 - 4) y = (1)(1 - 4) y = -
y = x 2/3^ (x 2 - 4); when x = 0 y = (0) 2/3^ ((0) 2 - 4) y = (0)(0 - 4) y = 0
y = x 2/3^ (x 2 - 4); when x = 1 y = (1) 2/3^ ((1) 2 - 4) y = (1)(1 - 4) y = -
The relative maximum would be at (0,0) and the relative minimum would be at (-1, -3) and (1, -3).
Exercise 5: Find the values of any relative extrema for the given function.
f(x) = x (4 - x 2 ) 1/
Find the derivative
f'(x) = x D (^) x(4 - x 2 ) 1/2^ + (4 - x 2 )1/2D (^) x(x)
f'(x) = x [
(4 - x 2 ) -1/2(-2x)] + (4 - x 2 ) 1/2(1)
f'(x) = -x 2 (4 - x 2 ) -1/2^ + (4 - x 2 )1/ f'(x) = (4 - x 2 ) -1/2^ [-x 2 + (4 - x 2 )] f'(x) = (4 - x 2 ) -1/2^ (-2x 2 + 4)
Find the critical numbers
0 = (4 - x 2 ) -1/2^ (-2x 2 + 4)
(4 - x 2 ) -1/2^ = 0
2
The fraction cannot be zero so we will set the denominator equal to zero.
4 - x 2 = 0 4 = x 2 ±2 = x
-2x 2 + 4 = 0 -2x 2 = - x 2 = 2
Evaluate the function at the critical numbers
f(x) = x (4 - x 2 ) 1/2; when x = - f(-2) = (-2)(4 - (-2) 2 ) 1/ = (-2)(0) 1/ = 0
