combinations

[mertz]

Hi. I'm trying to solve a couple of questions and I don't know if I'n doing them correctly or if I have the understanding to complete the work.

 

[daon]

Hi. I'm trying to solve a couple of questions and I don't know if I'n doing them correctly or if I have the understanding to complete the work.
h(x)=x^2-3x+12
g(x)=sqrt (x-6)

(gxh)(-2)
=g(x)h(x)
=sqrt(x-6) times x^2-3x+12
=sqrt ((-2)-6) times ((-2)^2-3(-2)+12)
=sqrt (-8) times (4+6+12)
=sqrt I-8I times 22
=62.22539674

is it "x-6" or is it x(-6) -- these are VERY different. You CANNOT plug -2 into this function! Supposing your work is correct, in any of the normal calculus classes i've taken or TA'd for, unless you are asked to estimate an answer, you should leave it in [simplified] symbolic form.

\(\displaystyle 22\sqrt{|-8|} = 44\sqrt{2}\)

(gxh)(x)
=g(x)+h(x)
=sqrt (x-6) +(x^2-3x+12)
stuck on where to go

Is it supposed to be (g+h)(x)? If so, you, again, cannot plug in -2. It is not an element in the domain of the given function g. Please clarify.

 

[mertz]

okay. thanks. i didn't even realize that i misstyped.

there's a couple of questions that deal with:
h(x)=x^2-3x+12
g(x)=sqrt x-6

1) (h o g)(10)
my understanding of that so far is:
=h(g(x))
=h(sqrt x-6)
=(sqrt x-6)^2-3(sqrt x-6)+12
etc...can't figure out how to solve that. i'm thinking factoring but i've been sitting around these questions trying to look for basic examples that will help me solve this myself and i'm not getitng anywhere.

after some calculations here's what i have so far for that one:
=(sqrt x-6)^2-3(sqrt x-6)+12
=(sqrt 10-6)^2-3(sqrt 10-6)+12
=(sqrt 4)^2-3(sqrt 4)+12
=2^2-3(2)+12
=4-6+12
=10

therefore (h o g) (10)=10

2) (gxh)(-2)
=g(x)h(x)
=sqrt(x-6) times x^2-3x+12
=sqrt ((-2)-6) times ((-2)^2-3(-2)+12)
=sqrt (-8) times (4+6+12)
=sqrt I-8I times 22
=62.22539674

oh okay so you're saying i should leave it in the radical form, because that's the appropriate notation and it should be 22 sqrt I-8I which = 22x 2sqrt 2 = 44sqrt 2

=sqrt I-8I times 22
=2srt 2 (22)
=44sqrt 2

3) (g+h)(x)
=g(x)+h(x)
=sqrt (x-6) +(x^2-3x+12)
stuck on where to go

4) (h/g)(x)

I was thinking that if i can get 2 & 3 done then I can get 1 and 4 done myself and move on with the rest of my work, but so far I'm not understanding this at all.

 

[daon]

okay. thanks. i didn't even realize that i misstyped.

there's a couple of questions that deal with:
h(x)=x^2-3x+12
g(x)=sqrt x-6

1) (h o g)(10)
my understanding of that so far is:
=h(g(x))
=h(sqrt x-6)
=(sqrt x-6)^2-3(sqrt x-6)+12
etc...can't figure out how to solve that. i'm thinking factoring but i've been sitting around these questions trying to look for basic examples that will help me solve this myself and i'm not getitng anywhere.

after some calculations here's what i have so far for that one:
=(sqrt x-6)^2-3(sqrt x-6)+12
=(sqrt 10-6)^2-3(sqrt 10-6)+12
=(sqrt 4)^2-3(sqrt 4)+12
=2^2-3(2)+12
=4-6+12
=10

therefore (h o g) (10)=10

Yes, that is correct.

2) (gxh)(-2)
=g(x)h(x)
=sqrt(x-6) times x^2-3x+12
=sqrt ((-2)-6) times ((-2)^2-3(-2)+12)
=sqrt (-8) times (4+6+12)
=sqrt I-8I times 22
=62.22539674

oh okay so you're saying i should leave it in the radical form, because that's the appropriate notation and it should be 22 sqrt I-8I which = 22x 2sqrt 2 = 44sqrt 2

=sqrt I-8I times 22
=2srt 2 (22)
=44sqrt 2

As written the answer is "undefined." As I stated in the last post, you cannot plug (-2) into \(\displaystyle g(x)=\sqrt{x-6}\), and hence you cannot evaluate \(\displaystyle (gh)(x)\). But you are correct on simplifying the radical.

3) (g+h)(x)
=g(x)+h(x)
=sqrt (x-6) +(x^2-3x+12)
stuck on where to go

If you are plugging in (-2) again, you can't do this. Algebraically, you also cannot "simplify" this at all. Same answer as above.

4) (h/g)(x)

I was thinking that if i can get 2 & 3 done then I can get 1 and 4 done myself and move on with the rest of my work, but so far I'm not understanding this at all.

Depends on what x's you are consdering.

First thing's first: find the domain of your function. If a question asks you to evaluate your function at a number not in its domain, your answer is "undefined" automatically, since the function has no definition at that value.

 

[mertz]

therefore (h o g) (10)=10

Yes, that is correct.

oh okay so you're saying i should leave it in the radical form, because that's the appropriate notation and it should be 22 sqrt I-8I which = 22x 2sqrt 2 = 44sqrt 2

=sqrt I-8I times 22
=2srt 2 (22)
=44sqrt 2

As written the answer is "undefined." As I stated in the last post, you cannot plug (-2) into \(\displaystyle g(x)=\sqrt{x-6}\), and hence you cannot evaluate \(\displaystyle (gh)(x)\). But you are correct on simplifying the radical.

[quote:v2l3crgq]
3) (g+h)(x)
=g(x)+h(x)
=sqrt (x-6) +(x^2-3x+12)
stuck on where to go

If you are plugging in (-2) again, you can't do this. Algebraically, you also cannot "simplify" this at all. Same answer as above.

4) (h/g)(x)

I was thinking that if i can get 2 & 3 done then I can get 1 and 4 done myself and move on with the rest of my work, but so far I'm not understanding this at all.

Depends on what x's you are consdering.

First thing's first: find the domain of your function. If a question asks you to evaluate your function at a number not in its domain, your answer is "undefined" automatically, since the function has no definition at that value.[/quote:v2l3crgq]

the way i understand it is that the domain is always X E R and then the range is where the limits are set.

the first one i think i just lucked out on because i've been watching some youtube videos to help me through it and i just gave it a try. hopefully i can replicate a similar varition of the same equation now that i know that it's (h of g of x).

for number 2 i understand why you're saying it's undefined because i cannot find the sqrt of a negative number...something like how my teacher was telling us that we cannot evaluate the log of a negative number. i'm going to leave the final answer as 44sqrt2 and add undefined so that i cover my bases.

for number 3, it's not asking me to plug in -2. i'm just supposed to solve for g(x)+h(x). my problem is that i keep looking at it and i think i have to factor something out and that i can't leave it as:

(g+h)(x)
=g(x)+h(x)
=sqrt x-6 + X^2-3x+12
=sqrt x-6 + x(x-3)+12

the part confusing me is sqrt x-6 because i keep thinking i have to change it into (x-6)^2 which is the same as sqrt x-6 and then just solve by expanding and collecting like terms.

for number 4...the division is freaking me out but here's what i have so far:

(h/g)(x)
=h(x)/g(x)
= (X^2-3x+12)/sqrt(x-6)

 

[mmm4444bot]

the way i understand it is that the domain is always X E R This is true for polynomial functions, like function h.

But this is wrong for function g. (The domain of the Square Root function does not include the negative numbers; graphically, function g is a horizontal shift of the square root function 6 units to the right, so the domain of g is clearly not all of the Reals.)

The domain of g(x) is all Real numbers that are 6 or more.



i'm going to leave the final answer as 44sqrt2 and add undefined so that i cover my bases.

That's silly. The answer cannot simultaneously be both a Real number and undefined.

In the Real number system (I assume you're in beginning calculus), the constant g(-2) is not defined. We can't do arithmetic with undefined numbers, so the expression 22 times g(-2) is not defined, either.

g(x) + h(x) cannot be simplified; leave it as the sum of a radical and a trinomial.

We can rationalize the denominator in h(x)/g(x), but that won't simplify the ratio because x - 6 is not a factor of x^2 - 3x + 12.

(I don't know if your teacher expects you to rationalize denominators in exercises like this one.)

So, I'd leave that one as is, too.

 

[mertz]

i stayed up all night and this morning and figured it out, but thank you for replying with your comments. i understand it a lot more than when i started out. i just need to take the time and think things through.