Would like some more knowledge about the product of functions

[mcastillo356]

TL;DR

Texbook don't seem to regard on the range and domains of function product. A search on the net gives clues, but I'd wish some more learning.

Hi, PF

I only have a clue on the topic I present; the answer involves the $\subset$, maybe $\subseteq$ concepts; I mean that the only answer I've obtained is that both range and domain of the product of two functions are the inclusion of both. I include a picture regarding the example of the textbook, which shows to functions between $0$ and $\pi$.

Greetings

 

[PeroK]

The domain of the product must be the intersection of the domains of the two functions, for the obvious reason.

The precise range of a product is more complicated, again for the obvious reason.

 

[weirdoguy]

domain of the product of two functions are the inclusion of both

What does that mean?

 

[mcastillo356]

Hi, PF, my question lacks of context and precision.
EXAMPLE 6 Find the area of the region bounded by $y=\Bigg(2+\sin{\displaystyle\frac{x}{2}}\Bigg)^2\cos{\displaystyle\frac{x}{2}}$, the $x$-axis, and the lines $x=0$ and $x=\pi$
Solution Because $y\leq 0$ when $0\leq x\leq \pi$, the required area is
$A=\displaystyle\int_0^{\pi}{\Bigg(2+\sin{\displaystyle\frac{x}{2}}\Bigg)^2\cos{\displaystyle\frac{x}{2}}}\,dx$
Let $v=2+\sin{\displaystyle\frac{x}{2}}$.
Then $dv=\displaystyle\frac{1}{2}\cos{\displaystyle\frac{x}{2}}\,dx$
$=2\displaystyle\int_2^3{\,v^2\,dv}=\displaystyle\frac{2}{3}\,v^3\Bigg|_2^3=\displaystyle\frac{2}{3}(27-8)=\displaystyle\frac{38}{3}$ square units

What does that mean?

Ups... Intersection. "The range of sum/difference/product/quotient of two functions is the intersection of the ranges of the individual functions" (quote from Stack Exchange).
Hope to have mended my previous post inconsistency. Greetings.

 

[Mark44]

EXAMPLE 6 Find the area of the region bounded by $y=\Bigg(2+\sin{\displaystyle\frac{x}{2}}\Bigg)^2\cos{\displaystyle\frac{x}{2}}$, the $x$-axis, and the lines $x=0$ and $x=\pi$
Solution Because $y\leq 0$ when $0\leq x\leq \pi$,

The inequality after "Because" isn't true. Perhaps you have the inequality reversed?
For the interval $[0, \pi]$, $\cos(x/2)$ and $\sin(x/2)$ both have values in the interval [0, 1].

 

[mcastillo356]

Perhaps you have the inequality reversed?

Yes.
$y\geq 0$ when $0\leq x\leq \pi$



Hi @Mark44, thanks a lot. Greetings!

 

[mcastillo356]

Hi, PF

I can draw
$y(x) = \begin{cases} y\geq 0 & \text{if }0\leq x \leq {\displaystyle\frac{\pi}{2}} \\ y\leq 0 & \text{if }\displaystyle\frac{\pi}{2} x \leq{\pi} \end{cases}$



This is easy, and doesn't add news to this thread

The question is: how should I manage to prove that $y\geq 0$ when $0\leq x\leq{\pi}$ for $$y=\Bigg(2+\sin{\displaystyle\frac{x}{2}}\Bigg)^2\cos{\displaystyle\frac{x}{2}}$$
avoiding the use of graphs?

The only mathematical evidence I have is that is true for the endpoints.

Greetings

 

[pasmith]

Hi, PF

I can draw
$y(x) = \begin{cases} y\geq 0 & \text{if }0\leq x \leq {\displaystyle\frac{\pi}{2}} \\ y\leq 0 & \text{if }\displaystyle\frac{\pi}{2} x \leq{\pi} \end{cases}$

View attachment 331364

This is easy, and doesn't add news to this thread

The question is: how should I manage to prove that $y\geq 0$ when $0\leq x\leq{\pi}$ for $$y=\Bigg(2+\sin{\displaystyle\frac{x}{2}}\Bigg)^2\cos{\displaystyle\frac{x}{2}}$$
avoiding the use of graphs?

The only mathematical evidence I have is that is true for the endpoints.

Greetings

You know that (2 + \sin (x/2))^2 > 0 for any x. So y can only be negative when \cos(x/2) < 0, which is not the case for x \in [0, \pi].