Maximum Value of Function and Integral of That Function

[Nazariy]

Hello friends,

Consider the following function:

\(\displaystyle \displaystyle{f(x)=\frac{x^6}{(x^2+1)^4}} \)

In the domain \(\displaystyle 0<x<1\), what is the maximum value of the function?

It is clearly 1/16, when x=1.

Now consider the following: \(\displaystyle \displaystyle{\int_0^1 \frac{x^6}{(x^2+1)^4}}\)

I cannot say that the integral of that is \(\displaystyle \leq \frac{1}{16}\), can I?

But the proof of what I am looking at is based on precisely this idea... hmm
[EDIT]: Or that \(\displaystyle \displaystyle{\int_0^1 \frac{x^6}{(x^2+1)^4} \leq \int_0^1 \frac{1}{16}}\)

 

[Ishuda]

Hello friends,

Consider the following function:

\(\displaystyle \displaystyle{f(x)=\frac{x^6}{(x^2+1)^4}} \)

In the domain \(\displaystyle 0<x<1\), what is the maximum value of the function?

It is clearly 1/16, when x=1.

Now consider the following: \(\displaystyle \displaystyle{\int_0^1 \frac{x^6}{(x^2+1)^4}}\)

I cannot say that the integral of that is \(\displaystyle \leq \frac{1}{16}\), can I?

But the proof of what I am looking at is based on precisely this idea... hmm
[EDIT]: Or that \(\displaystyle \displaystyle{\int_0^1 \frac{x^6}{(x^2+1)^4} \leq \int_0^1 \frac{1}{16}}\)

Why can't you say that? Some speech impediment? Yes, you can say that since the function is continuous and integrable on [0, 1].

EDIT: Actually you don't even need the continuous part of it.

 

[Nazariy]

Why can't you say that? Some speech impediment? Yes, you can say that since the function is continuous and integrable on [0, 1].

Ahhhhhhh

I have just imagined y=x in the domain 0<x<1, and how max value is 1 at x=1. And the integral of y=x in that domain is 1/2... and how the largest area under curve can only be 1 if the curve gets pushed up... ok....

 

[Nazariy]

Why can't you say that? Some speech impediment? Yes, you can say that since the function is continuous and integrable on [0, 1].

EDIT: Actually you don't even need the continuous part of it.

Thank you, Ishuda, as always