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Exercise Set 12: Finding Open Intervals, Relative Extrema and Critical Points of Functions, Study notes of Calculus

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.

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2021/2022

Uploaded on 09/27/2022

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Review Exercise Set 12
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 = x3 + x2 - 8x + 5
pf3
pf4
pf5
pf8
pf9

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Download Exercise Set 12: Finding Open Intervals, Relative Extrema and Critical Points of Functions and more Study notes Calculus in PDF only on Docsity!

Review Exercise Set 12

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)

Review Exercise Set 12 Answer Key

Exercise 1: Find the open intervals on the graph where the given function is increasing or decreasing.

Increasing: ( −2, 0)

Decreasing: ( −∞ −, 2 ) ∪ ( 0,∞)

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

x x x

x

x

f x

Increasing: ( )

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

x

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.

3 x = 0

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

4 − x

The fraction cannot be zero so we will set the denominator equal to zero.

4 − x^2 = 0

4 - x 2 = 0 4 = x 2 ±2 = x

-2x 2 + 4 = 0 -2x 2 = - x 2 = 2

x = ± 2

The critical numbers are -2, - 2 , 2 , and 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