# Serge Lang: Increasing/Decreasing Function

• Cosmophile
In summary: Can you show that f1 in prob. 11a is increasing? It looks like Lang's intent is for you to work through the various subparts of this problem before tackling the inequality in your first post.BTW, this is definitely leading up to the Maclaurin series for sin(x) and cos(x).
Cosmophile

## Homework Statement

Consider only values of ##x \geq 0##, and let ##f(x) = -x + \frac {x^3}{3!} + \sin x##. Show whether ##f## is increasing or decreasing.

## Homework Equations

$$f(x) = -x + \frac {x^3}{3!} + \sin x$$
$$f'(x) = -1 + \frac {x^2}{2} + \cos x$$

## The Attempt at a Solution

I know that ##f## is increasing whe ##f' > 0##, and that ##f## is decreasing when ##f' < 0##. In the first case, I have $$f'(x) = -1 + \frac {x^2}{2} + \cos x > 0$$
$$x^2 + 2 \cos x > 2$$
$$|x| > 2|(1- \cos x)|$$

Unfortunately, I'm having a hard time making any sense of this solution. For some reason, trigonometric functions are the only functions I've dealt with that give me any issue with these types of problems. It's demoralizing, really. Anyway, any help is greatly appreciated, as always!

Cosmophile said:

## Homework Statement

Consider only values of ##x \geq 0##, and let ##f(x) = -x + \frac {x^3}{3!} + \sin x##. Show whether ##f## is increasing or decreasing.

## Homework Equations

$$f(x) = -x + \frac {x^3}{3!} + \sin x$$
$$f'(x) = -1 + \frac {x^2}{2} + \cos x$$

## The Attempt at a Solution

I know that ##f## is increasing whe ##f' > 0##, and that ##f## is decreasing when ##f' < 0##. In the first case, I have $$f'(x) = -1 + \frac {x^2}{2} + \cos x > 0$$
$$x^2 + 2 \cos x > 2$$
$$|x| > 2|(1- \cos x)|$$

Unfortunately, I'm having a hard time making any sense of this solution. For some reason, trigonometric functions are the only functions I've dealt with that give me any issue with these types of problems. It's demoralizing, really. Anyway, any help is greatly appreciated, as always!
I don't see that approach as bearing fruit. Have you studied Maclaurin and Taylor series, yet? This problem seems to be aimed in that direction.
Here's the Maclaurin series for sin(x):
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots + (-1)^n\frac{x^{2n + 1}}{(2n + 1)!} + \dots$$

Mark44 said:
I don't see that approach as bearing fruit. Have you studied Maclaurin and Taylor series, yet? This problem seems to be aimed in that direction.
Here's the Maclaurin series for sin(x):
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -+ \dots + (-1)^n\frac{x^{2n + 1}}{(2n + 1)!} + \dots$$

I've seen them, but I haven't studied them. I think Lang may be hinting to them for future use, though. Here is the full problem (problem number 11):

Can you show that f1 in prob. 11a is increasing? It looks like Lang's intent is for you to work through the various subparts of this problem before tackling the inequality in your first post.

BTW, this is definitely leading up to the Maclaurin series for sin(x) and cos(x).

Cosmophile said:

## Homework Statement

Consider only values of ##x \geq 0##, and let ##f(x) = -x + \frac {x^3}{3!} + \sin x##. Show whether ##f## is increasing or decreasing.

## Homework Equations

$$f(x) = -x + \frac {x^3}{3!} + \sin x$$
$$f'(x) = -1 + \frac {x^2}{2} + \cos x$$

## The Attempt at a Solution

I know that ##f## is increasing whe ##f' > 0##, and that ##f## is decreasing when ##f' < 0##. In the first case, I have $$f'(x) = -1 + \frac {x^2}{2} + \cos x > 0$$
$$x^2 + 2 \cos x > 2$$
$$|x| > 2|(1- \cos x)|$$

Unfortunately, I'm having a hard time making any sense of this solution. For some reason, trigonometric functions are the only functions I've dealt with that give me any issue with these types of problems. It's demoralizing, really. Anyway, any help is greatly appreciated, as always!

You should first try to determine the behavior of ##f(x)## for ##0 \leq x \leq 2 \pi##. For ##x > 2 \pi## the polynomial part continues to increase, while the ##\sin## part just repeats its behavior on the initial segment.

Mark44 said:
Can you show that f1 in prob. 11a is increasing? It looks like Lang's intent is for you to work through the various subparts of this problem before tackling the inequality in your first post.

BTW, this is definitely leading up to the Maclaurin series for sin(x) and cos(x).

$$f(x) = x - \sin x$$
$$f'(x) = 1 - \cos x$$

##f## is increasing when ##f'>0##, so we have ##1- \cos x > 0 \to 1 \geq \cos x##, which is true ##\forall x \in \mathbb R## because the range of ##\cos x## is ##[-1,1]##.

This is the same method I used on the problem I posted, but the outcome was not nearly as friendly (and also is almost definitely wrong).

Ray Vickson said:
You should first try to determine the behavior of ##f(x)## for ##0 \leq x \leq 2 \pi##. For ##x > 2 \pi## the polynomial part continues to increase, while the ##\sin## part just repeats its behavior on the initial segment.

I'm working on establishing that level of intuition; my schooling has always been algorithmic. What tips do you have for doing this? Thanks!

Cosmophile said:
I'm working on establishing that level of intuition; my schooling has always been algorithmic. What tips do you have for doing this? Thanks!
I would suggest looking at the graphs of simple polynomials, such as y = x, y = x2, y = x3, and so on, to get a visual feel (as opposed to only algebraic, which is how I interpret your use of "algorithmic") for how these functions behave, both near the origin as well as farther away.

For polynomials with more terms, satisfy yourself that, for large x, the dominant term of y = x3 - 2x2 + 3 (for example) is the x3 term. The larger x is, the less influence the other terms have on the behavior.

Cosmophile
Cosmophile said:
I'm working on establishing that level of intuition; my schooling has always been algorithmic. What tips do you have for doing this? Thanks!

Well, if ##p(x) = x^3/3! - x = x^3/6 - x##, then ##p'(x) = x^2 / 2 - 1 > 0## if ##x > \sqrt{2}##. So, since ##\pi/2 > \sqrt{2}##, the function ##p(x)## is strictly increasing for ##x > \pi/2##, which is even stronger than saying for ##x > 2 \pi##. As for the ##\sin## part repeating, that just follows from the fact that the trig functions are periodic, with period ##2 \pi##.

I agree with Mark44, that drawing graphs can be helpful. Looking at a graph may not constitute an actual proof, but it helps you to gain insight and to (sometimes, at least) see what type of "analytical" reasoning might be best.

## 1. What is a "Increasing/Decreasing Function" in math?

An increasing/decreasing function is a type of mathematical function that describes a relationship between two variables, where one variable increases or decreases while the other variable changes. In an increasing function, the output value increases as the input value increases, while in a decreasing function, the output value decreases as the input value increases.

## 2. How can I determine if a function is increasing or decreasing?

To determine if a function is increasing or decreasing, you can plot the function on a graph and observe the trend of the curve. If the curve is going upwards from left to right, the function is increasing, and if the curve is going downwards from left to right, the function is decreasing. You can also take the first derivative of the function and analyze its sign to determine if it is increasing or decreasing.

## 3. Are there any specific rules for finding increasing/decreasing functions?

Yes, there are some rules that can help you identify increasing/decreasing functions. For example, if the first derivative of a function is positive for all values of the independent variable, then the function is increasing. Similarly, if the first derivative is negative for all values of the independent variable, then the function is decreasing. Additionally, a function can also be increasing or decreasing over a specific interval, depending on the behavior of its first derivative in that interval.

## 4. What is the significance of knowing if a function is increasing or decreasing?

Knowing if a function is increasing or decreasing is important because it helps in understanding the behavior of the function and making predictions about its future values. It also allows us to determine the maximum and minimum values of the function, which can be useful in optimization problems. Moreover, the concept of increasing/decreasing functions is widely used in economics, physics, and other fields to model and analyze various real-world situations.

## 5. Can a function be both increasing and decreasing?

No, a function cannot be both increasing and decreasing at the same time. This is because if a function is increasing, it means that its output value is increasing as the input value increases, while in a decreasing function, the output value decreases as the input value increases. These two behaviors are mutually exclusive, so a function can only be either increasing or decreasing.

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