Sample Problems on Increasing and Decreasing Functions
Problem 1: Given the function g(x) = 3x2 – 12, find the intervals on -3 < x < 3 where g(x) is increasing and decreasing.
Solution:
Given function: g(x) = 3x2 – 12
Differentiate w.r.t. x, we get
g'(x) = 6x
For increasing and decreasing
Put g'(x) = 0
g'(x) = 6x = 0
So, x = 0
Intervals: (-3, 0), (0, 3)
At x = -2, g'(-2) = -12 < 0
At x = 2, g'(2) = 12 > 0
So, for -3 < x < 3, g(x) is decreasing on (-3, 0) and increasing on (0, 3).
Problem 2: Given the derivative of f(x), f'(x) = -10x2 + 40x, find the intervals where f(x) is increasing and decreasing.
Solution:
Given: f'(x) = -10x2 + 40x
For increasing and decreasing
Put f'(x) = 0
f'(x) = -10x2 + 40x = 0
So, x = 4, 0
Intervals: (−∞, 0), (0, 4), (4, ∞)
So, at x = -1, f'(-1) = -50 < 0
at x = 1, f'(1) = 30 > 0
at x = 5, f'(5) = -50 < 0
So, f(x) is increasing on (0, 4) and decreasing on (−∞, 0), (4, ∞)
Problem 3: Given the function g(x) = 5x2 – 20x + 100, find the intervals where g(x) is increasing and decreasing.
Solution:
Given: g(x) = 5x2 – 20x + 100
Differentiate w.r.t. x, we get
g'(x) = 10x – 20
For increasing and decreasing
Put g'(x) = 0
g'(x) = 10x – 20 = 0
x = 2
Intervals: (−∞, 2), (2, ∞)
At, x = 1, g'(1) = -10 < 0
At x = 3, g'(3) = 10 > 0
So, g(x) is decreasing on (-∞, 2) and increasing on (2, ∞)
Problem 4: Given the function s(x) = 6x3 – x2, find the intervals on 0 < x < 10 where s(x) is increasing and decreasing.
Solution:
Given: s(x) = 6x3 – x2
Differentiate w.r.t. x, we get
s'(x) = 18x2 – 2x
For increasing and decreasing
Put s'(x) = 0
s'(x) = 18x2 – 2x = 0
x = 1/9, -1/9
Intervals: (0, 1/9)
Here, -1/9 is not in the given interval, 0 < x < 10
So, g'(1/10) = -0.02 < 0
Hence, for 0 < x < 10, g(x) is decreasing on (0, 1/9)
Problem 5: Given g'(x) = 7x2 – 8, find the intervals where g(x) is increasing and decreasing.
Solution:
Given: g'(x) = 7x2 – 8
For increasing and decreasing
Put g'(x) = 0
g'(x) = 7x2 – 8 = 0
x = {√(8/7), -√(8/7)}
Intervals: (-∞,√(8/7)), (-√(8/7), √(8/7)), (√(8/7), ∞)
So, At x = -10, g'(-10) = 692 > 0
At x = 0, g'(0) = -8 < 0
At x = 10, g'(10) = 692 < 0
Hence, g(x) is increasing on (-∞,√(8/7))and (√(8/7), ∞), and decreasing on (-√(8/7), √(8/7))
Increasing and Decreasing Functions
If you’re studying calculus, then you’re probably familiar with the concepts of increasing and decreasing functions. These terms refer to the behavior of a function as its input values change. Specifically, an increasing function is one that becomes larger as its input values increase, while a decreasing function is one that becomes smaller as its input values increase. Understanding these concepts is crucial for solving a variety of calculus problems, from finding maximum and minimum values to understanding the behavior of graphs.
In this article, we’ll delve deeper into increasing and decreasing functions, exploring how to identify them, and how to use them to solve problems in calculus.
Table of Content
- Increasing Function Definition
- Decreasing Function Definition
- Constant Function Definition
- Rules to Check Increasing and Decreasing Functions
- Graph of Increasing, Decreasing, and Constant Function
- Properties of Increasing & Decreasing Functions
- How to Find Increasing and Decreasing Intervals
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