# Would like some more knowledge about the product of functions

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• mcastillo356
mcastillo356
Gold Member
TL;DR Summary
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

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.

mcastillo356
mcastillo356 said:
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
weirdoguy said:
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.

mcastillo356 said:
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].

Mark44 said:
Perhaps you have the inequality reversed?
Yes.
##y\geq 0## when ##0\leq x\leq \pi##

Hi @Mark44, thanks a lot. Greetings!

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}##

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

mcastillo356 said:
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

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 $x \in [0, \pi]$.

mcastillo356

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