Functions (1 Viewer)

[cherry1991]

Let f(x) = -2 on x squared

a) Find Lim f(x)
x --> +/- infinity

b) Find lim f(x)
x-->0

Find the domain and range is it Domain: all x, x cannot = 0 and Range all y, y< 1 and cannot equal 0?

d) can f(x) = 0

its really confusing me

 

[DownInFlames]

Let f(x) = -2 on x squared

a) Find Lim f(x)
x --> +/- infinity

b) Find lim f(x)
x-->0

Find the domain and range is it Domain: all x, x cannot = 0 and Range all y, y< 1 and cannot equal 0?

d) can f(x) = 0

its really confusing me

please correct me if I'm wildly wrong. Am no longer in any way confident in my maths theory.


a)
as x -> +infinity, the denominator gets really big/positive, so f(x) will become really small and negative --- [f(x) -> -0]

as x -> -infinity, your denominator will do the same as above, because x^2 can't be negative. so again ----[f(x) -> -0]

b)
as x -> 0, f(x) -> -infinity.


I think.

f(x) cannot equal 0 because the numerator is not 0, the denominator cannot be (would be no solution)

 

[tommykins]

Let f(x) = -2 on x squared

a) Find Lim f(x)
x --> +/- infinity

y approaches 0.

b) Find lim f(x)
x-->0

bah minds blank on what this is asking =\

Find the domain and range is it Domain

Domain: all x =/= 0, range all y < 0

d) can f(x) = 0

no

 

[tacogym27101990]

Let f(x) = -2 on x squared

a) Find Lim f(x)
x --> +/- infinity

b) Find lim f(x)
x-->0

Find the domain and range is it Domain: all x, x cannot = 0 and Range all y, y< 1 and cannot equal 0?

d) can f(x) = 0

its really confusing me

a) i'd say as x-->+/-infinity, y--> 0^-
just divide throughout by the highest power (in this case x^2)
and then remember that 1/x=0

b) x-->0, y -->-infinity

c) Domain: all real x except x=0. Range: all real y<0

d)y cannot = 0


sorry if its wrong, and for the lack of explaination...

too late for this shit

sorry just replace the y's with f(x)

 

[cherry1991]

Thankyou so much u guys =]

 

[Sew2289]

Not to hijack the thread or anything, but over the holidays my brain has stopped working. Can some please answer:

f(x)=2x^2-4x

Find F (x-1)

Thanks.

 

[lyounamu]

Not to hijack the thread or anything, but over the holidays my brain has stopped working. Can some please answer:

f(x)=2x^2-4x

Find F (x-1)

Thanks.

In place of x, you put (x-1)

i.e. f(x-1) = 2(x-1)^2 - 4(x-1)
= 2(x^2 -2x +1) - 4x +4
= 2x^2 -4x +2 -4x +4
= 2x^2 -8x +6

 

[Sew2289]

Thanks for the quick response mate. It's all coming back now!

 

[lyounamu]

Thanks for the quick response mate. It's all coming back now!

Glad to hear!

Doing some revision will quickly get rid of that dilemma!

 

[adidasboy]

How do you do this?
If f (x)= x^3 - 2, find
[f(x)]^2

thanks

 

[lyounamu]

How do you do this?
If f (x)= x^3 - 2, find
[f(x)]^2

thanks

That is essentially (x^3 -2)^2
Therefore, (x^3 - 2)^2 = x^6 - 4x^3 +4

 

[adidasboy]

Awesome, thanks for answering.

How do i answer the question
Show that f(x) is odd where f(x) = x.

I thought odd is f(-x)= -f(x).

sorry for these stupid questions btw. Revising my maths lol ^_^

 

[lyounamu]

Awesome, thanks for answering.

How do i answer the question
Show that f(x) is odd where f(x) = x.

I thought odd is f(-x)= -f(x).

sorry for these stupid questions btw. Revising my maths lol ^_^

When you get the question like that all you have to do is the followings:
f(x) = x
Therefore f(-x) = -x
= - f(x)
Since f(-x) = -f(x), it is an odd function.

In summary, write the f(x) first. Then write then f(-x) in the second line.
On the third line, prove that f(-x) is equal to either f(x) or -f(x). If f(-x) is not equal to neither f(x) or -f(x), you can say that the function is neither an odd nor an even function.

 

[x.Exhaust.x]

Awesome, thanks for answering.

How do i answer the question
Show that f(x) is odd where f(x) = x.

I thought odd is f(-x)= -f(x).

sorry for these stupid questions btw. Revising my maths lol ^_^

f(x) = f(-x) is Even [Rule 1]
f(-x) = -f(x) is Odd [Rule 2]

When f(x) = x

f(-x) = -x
-f(x) = -x

Both are negative, therefore, odd [Rule 2].

Edit: I didn't see your answer lyounamu. Meh. Ask more if you need more help adidasboy.

 

[adidasboy]

Are these functions even, odd or neither?
A) y = x^3 / x^5-x^2
B) y = x-3 / x+3

What happens when there is a y instead of the f(x) ?

 

[lyounamu]

Are these functions even, odd or neither?
A) y = x^3 / x^5-x^2
B) y = x-3 / x+3

What happens when there is a y instead of the f(x) ?

In this question you regard y = f(x). f(x) means there is only one value of y for every value of x.

In that kind of question, you work out like the followings:
For even functions, f(-x) = f(x)
For odd functions, f(-x) = -f(x)

Let f(x) = x^3/x^5-x^2
Then f(-x) = (-x)^3/(-x)^5-(-x)^2
= -x^3/-x^5-x^2

From the above rules, we can see that f(-x) is not equal to f(x) nor -f(x).
Therefore, the function is neither odd nor even function.

For the second question, you work out similarly like the followings:
For even functions, f(-x) = f(x)
For odd functions, f(-x) = -f(x)

Let f(x) = x-3 / x+3
Then f(-x) = -x-3 / -x +3
= -(x+3)/-(x-3)
= x+3/x-3
From the above rules, we can see that f(-x) is not equal to f(x) nor -f(x). Therefore, the function is neither odd nor even function.

 

[adidasboy]

Find Domain and Range of these functions.
y= square root of x^2+2x-3

D:?
R: y>_ 0

How do I work out the Domain?
Thanks

 

[adidasboy]

Was the range (y>_ 0) given to you?

If so, write your working like the followings:

y=x^2+2x-3
Since y>_0, x^2+2x-3 >_ 0
x^2 + 2x-3 >_ 0
(x+3)(x-1) >_0
From here, you have to either test the value or draw the graph to see which x value satisifes the equation.
I drew the graph but I cannot draw here.
(After you either test it or draw the graph,) Therefore x<_-3 or x>_1.

Nah, but you know from looking at it. Thank you for your time. So kind helping me with this ^_^

Oh so you ignore the square root and work it out.

 

[adidasboy]

But when y = square root of x
y is always y>_0 because when you square root something it always is above/equal to 0

i think i was unclear in my question

i added in the y>_0
wasn't given

 

[Continuum]

Find Domain and Range of these functions.
y= square root of x^2+2x-3

D:?
R: y>_ 0

How do I work out the Domain?
Thanks

f(x) = sqrt(x^2+2x-3)
Anything inside a square root sign must be greater than or equal to zero. Hence:
x^2+2x-3 >= 0
(x+3)(x-1) >= 0
x <= 3 and x >= 1