Maths: Functions...

demon8991 · 25 Oct 2006 at 16:49

f(x) = square root of (9-x2)

Is this the natural domain:

x belongs to the real numbers with x < or = to 3)

and the range:

[0,3]

Also is

f(x) = sinx + tanx

a periodic (i think it is but im not sure what the period is 2pi maybe?)

Thanks guys

Dave · 25 Oct 2006 at 16:55

demon8991 said:
f(x) = square root of (9-x2)

Is this the natural domain:

x belongs to the real numbers with x < or = to 3)

and the range:

[0,3]

Yep, the domain of f(x) for the real number system is x<=3. For the natural system it's 0 < x =< 3

Also is

f(x) = sinx + tanx

a periodic (i think it is but im not sure what the period is 2pi maybe?)

Thanks guys

Yep, except the period is pi.

demon8991 · 25 Oct 2006 at 16:57

Thanks mate

Dave said:
Yep, the domain of f(x) for the real number system is x<=3.

Is the range right too?

demon8991 · 25 Oct 2006 at 17:01

Dave said:
Yep, except the period is pi.

The only reason i thought it was 2pi was becuase sinx has a period of 2pi so i couldnt get my head round why sinx + tanx would be pi?

Dave · 25 Oct 2006 at 17:04

demon8991 said:
Thanks mate

Is the range right too?

Sorry, didn't check.

Range:

Min is at +- 3, where f(x) = sqrt(9 - (3)^2) = 0
Max is at 0, where f(x) = sqrt(9 - (0)^2) = 3

So yup, range is 0 =< f(x) =< 3

Here's the graph of it: (not sure why it cuts off just before 3 on my VirtualTI, on my TI83 it graphs it)

demon8991 said:
The only reason i thought it was 2pi was becuase sinx has a period of 2pi so i couldnt get my head round why sinx + tanx would be pi?

Because tan(x) has a period of pi. There is an algebraic way of solving it but I've completely forgotten (something like let f(x+p) = f(x)) so here's the graph: (each asymptote is pi away from each other)

demon8991 · 25 Oct 2006 at 17:09

Thanks mate you have helped me no end.

That little graph drawing thing is well good.

Dave · 25 Oct 2006 at 17:12

If it's for natural numbers then I messed up the domain sorry - if its for the natural system it's 0 < x =< 3, and that'd make the range 0 < f(x) =< 3. (as 0 is not a natural number)

Then again if it's only naturals then we can shorten it further as x = {1, 2, 3} and thus 0 <= f(x) <= 2sqrt(2) for the range, as f(x) = {2sqrt(2), sqrt(5), 0}

Sorry, missed the whole natural number part

demon8991 · 25 Oct 2006 at 17:23

Also another question im unsure about.

is f(x) = x4 + xsinx an odd, even or neither function?

Psiko · 25 Oct 2006 at 18:06

demon8991 said:
Also another question im unsure about.

is f(x) = x4 + xsinx an odd, even or neither function?

What does f(-x) equal?

demon8991 · 25 Oct 2006 at 18:39

Im really not sure about the question ive got a feeling its an even function but i really am not sure.

Psiko · 25 Oct 2006 at 18:44

demon8991 said:
Im really not sure about the question ive got a feeling its an even function but i really am not sure.

What I'm saying is put -x into f.

f(-x) = (-x)^4 + (-x) sin(-x)

What does that equal?? Do you know the definition of odd and even functions?
Even means f(-x) = f(x), odd means f(-x) = -f(x).

Dave · 25 Oct 2006 at 18:48

f(x) = x^4 + xsin(x)

A function is even if f(x) = f(-x), and odd if f(-x) = -f(x)

f(-x) = (-x)^4 + (-x)sin(-x)

Now, sin(-x) = -sin(x) and (-x)^4 = (x^2)^2 = x^4;

f(-x) = x^4 + sin(x)

Thus f(-x)=f(x) and the function is even.

Quick graph shows it's even too as it's symmetric about the y axis:

demon8991 · 25 Oct 2006 at 19:25

Thanks guys you have helped no end.