Functions Help needed!!

[LukeGernon]

Need help guys, hopefully someone can show me how to do this. Its the (not equal) sign which is throwing me off I can put the function together. Just don't know where to start with it all. If someone could write it out for me with an explanation to help me understand ild appreciate it. HND student studying engineering btw

MUCH THANKS IN ADVANCE!

 

[Otis]

… I can put the [functions] together …

Hi Luke. Can you show us what you got in part (a), when you composed the functions?

Part (a) tells us to use the output from function h as the input to function g. [By the way, their expression g(h(x) is missing a close-parenthesis.]

So, what did you get, when you replaced symbol x in g(x) with the expression 5x+2?

The statement x ≠ 3 is a domain statement, for function g. It means the domain is all Real numbers except 3. The exercise doesn't ask for the domains of g(h(x)) and h(g(x)), but it's good practice to consider them, anyway.

Here's a link to the posting guidelines summary.

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[LukeGernon]

here is what I got first attempt and then sat there confused on google for 2 hours + lol

 

[Otis]

That's a good start, Luke. Next, simplify the new denominator.

The last line of your work contains two misstatements.

(1) Write g(h(x))= instead of g=

(2) Don't write ≠3

Just like an equals sign, the symbol ≠ requires an expression on each side of it. Also, the domain statement x≠3 is for function g only. The domain for the composed function g(h(x)) is different.

Do you understand why x cannot be 3, in function g? If you do, then use the same logic to determine the domain for function g(h(x)). If you're not sure, please say so.

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[LukeGernon]

So b

That's a good start, Luke. Next, simplify the new denominator.

The last line of your work contains two misstatements.

(1) Write g(h(x))= instead of g=

(2) Don't write ≠3

Just like an equals sign, the symbol ≠ requires an expression on each side of it. Also, the domain statement x≠3 is for function g only. The domain for the composed function g(h(x)) is different.

Do you understand why x cannot be 3, in function g? If you do, then use the same logic to determine the domain for function g(h(x)). If you're not sure, please say so.

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So by using this ≠3 are they saying its impossible for g to equal 3. and I'm still confused as to where it goes lol how do we get rid of it for the equation or have I just put it in the incorrect place?

thank you so much btw, promise I'm not silly just confused.

 

[Otis]

Hi Luke. Please don't write ≠3 to mean x≠3. Write x≠3, instead.

… by using this ≠3 are they saying its impossible for g to equal 3[?] …

No -- they're not talking about function g's outputs. It's a domain statement. A function's domain is the set of numbers which may be used as inputs to the function.

… have I just put it in the incorrect place? …

Yes. The domain statement x≠3 belongs to function g only.

For function g, the domain statement x≠3 tells us that any Real number will work as an input to function g except the number 3. Here's what happens, if we try to evaluate the specific output g(3).

g(x) = 6/(3 - x)

g(3) = 6/(3 - 3) = 6/0

Division by zero is not defined. Therefore, the number g(3) does not exist. That is why x≠3, for function g.

Because they haven't asked us to state domains for the composed functions, we could ignore the given domain statement for function g. It's not needed, to work out the composed functions.

They've asked you to find g(h(x)) and h(g(x)) only. So far, you've found g(h(x)) correctly, but you didn't simplify your result. Can you do that now? We can discuss domains further, after you finish the exercise.

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[Steven G]

You can't compute g(3) as this will involve dividing by 0. What this means is that you can't put 3 inside the bold parenthesis in g( ). I hope that is clear, again if you do you will end up diving by 0--and you can't divide by 0.

If we were computing g(x), then x could not be 3. If we were computing g(x+7), then x+7 could not be 3. If we were computing g(anything), then anything could not be 3. In our case we are computing g(f(x)) so f(x) can't be 3. That is 5x+2 can't be 3 or 5x can't be 1 or x can't be 1/5. Believe it or not after what you wrote for g(f(x)) you should then say that x can't be 1/5. Is this clear? Again, regardless of what you are taking g of, it can't be 3!

 

[LukeGernon]

Hi Luke. Please don't write ≠3 to mean x≠3. Write x≠3, instead.


No -- they're not talking about function g's outputs. It's a domain statement. A function's domain is the set of numbers which may be used as inputs to the function.


Yes. The domain statement x≠3 belongs to function g only.

For function g, the domain statement x≠3 tells us that any Real number will work as an input to function g except the number 3. Here's what happens, if we try to evaluate the specific output g(3).

g(x) = 6/(3 - x)

g(3) = 6/(3 - 3) = 6/0

Division by zero is not defined. Therefore, the number g(3) does not exist. That is why x≠3, for function g.

Because they haven't asked us to state domains for the composed functions, we could ignore the given domain statement for function g. It's not needed, to work out the composed functions.

They've asked you to find g(h(x)) and h(g(x)) only. So far, you've found g(h(x)) correctly, but you didn't simplify your result. Can you do that now? We can discuss domains further, after you finish the exercise.

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think ive done it correctly now, Understanding that we can't use x = 3 because within the equation theres a 3 - which would result in zero.

 

[Steven G]

 

[Steven G]

Let's see if you are correct. g(h(3)) = g(17) = 6/(3-17) = -6/14 = -3/7. So what is the problem with x not being 3.

What about when x = 1/5?

g(h(1/5)) = g( 3 ) = 6/(3-3) = 6/0 which is undefined. Oh, so x can't be 1/5.

Now please go back and read post #7 to see where I got 1/5 from.

 

[LukeGernon]

Let's see if you are correct. g(h(3)) = g(17) = 6/(3-17) = -6/14 = -3/7. So what is the problem with x not being 3.

What about when x = 1/5?

g(h(1/5)) = g( 3 ) = 6/(3-3) = 6/0 which is undefined. Oh, so x can't be 1/5.

Now please go back and read post #7 to see where I got 1/5 from.

Feel like I understand what you've said in post 7 but your last post has just confused me loads lol, pls help

 

[LukeGernon]

I worked through the second question factorising and simplifying, this look at all okay?

 

[Steven G]

Feel like I understand what you've said in post 7 but your last post has just confused me loads lol, pls help

My last post shows that x=3 is a valid number for an input while x=1/5 is not a valid input. If you understood my previous post then why did you say x can't be 3 instead of x can't be 1/5.
Please review that post again and see why it is that x can't be 1/5.

 

[Steven G]

I worked through the second question factorising and simplifying, this look at all okay?

When you divided by 5 you made a careless mistake. Also, please do not use x for multiplication and for the variable x, please!

 

[Otis]

think ive done [part (a)] correctly now, Understanding that we can't use x = 3

Hi Luke. Your latest result is correct, but your work still contains misstatements. I've told you twice that x≠3 does not apply to part (a). I don't understand why you continue to associate x≠3 with the function in part (a).

The answer to part (a) is

g(h(x)) = 6/(1 - 5x)

That's all.

(We can discuss domains, after you finish the exercises. Those two exercises do not ask about domains.)

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Here are issues with your latest work on part (a).



Do not write an equals sign between a result and your scratch work. (The equation shown above is true only when x=-1.) We haven't been provided with any values for x; therefore, we're not allowed to say those two sides are equal.




The letter 'g' does not represent a number.; it's just a name (for one of the given functions).

The correct symbol for the number you're talking about is g(h(x)).

Don't include ≠3. That has nothing to do with part (a).

What's up with the X that you've added, at the right end above? Is it meant to tell readers to ignore the entire line? If so, there's a better way to do that.

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[Otis]

I worked through the second question factorising and simplifying, this look at all okay?

No -- the first line is not correct. The rest of your steps seem okay, but your final result is incorrect because you started with the wrong equation.

In part (b), we use g(x) as the input to function h, so we replace the input symbol x in function h with the expression 6/(3-x):

h(g(x)) = 5*[6/(3 - x)] + 2

Simplify the right-hand side, above.

---

I'd noticed that Jomo asked you to not use 'x' as a multiplication symbol, anymore. That's important. After starting algebra, we use other ways to show multiplication. (Scientific notation is one exception.)

You may type an asterisk, instead, as I've shown (highlighted in red) above. I typed that asterisk, as an example. We don't really need to type a multiplication operator, in that case, because the grouping symbols around 6/(3-x) are sufficient to indicate the multiplication by 5.

You may use a centered dot, like this: a·b

You may use grouping symbols, like this: a(b).

By the way, did you create the content in the post #1 image, or is that an image of what you were given by your instructor?

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[Steven G]

As other helper know, I am a fanatic when it comes to equality sign.

In my opinion, just stating what g( h(x) ) equals is not enough. I feel that you should state the restriction on x, which is x is not 1/5

 

[LukeGernon]

I believe this is correct, Please check ive put the correct function composition lol, I sat here doing examples to make sure it was the correct way round I think there may be a typo/mistake on one of these replies or I am still not getting it ?

thank you guys for your patience.

 

[Dr.Peterson]

I believe this is correct, Please check ive put the correct function composition lol, I sat here doing examples to make sure it was the correct way round I think there may be a typo/mistake on one of these replies or I am still not getting it ?

thank you guys for your patience.

At the top, you wrote the correct expressions; but then for (a) you said that g(h(x)) equals the expression for h(g(x)). Did you just stop thinking for a moment? So what you've really done is part (b).

But you have correctly simplified that expression for h(g(x)). Others will point out the dangerous use of the multiplication sign (I'd use a dot or parentheses, like 6(5), for clarity), but I'd rather focus on what you've done correctly with regard to the core of the work.

To be complete, you should include an explicit statement of the domain.

Now you need to do (a), which is a little harder in my opinion.

 

[LukeGernon]

At the top, you wrote the correct expressions; but then for (a) you said that g(h(x)) equals the expression for h(g(x)). Did you just stop thinking for a moment? So what you've really done is part (b).

But you have correctly simplified that expression for h(g(x)). Others will point out the dangerous use of the multiplication sign (I'd use a dot or parentheses, like 6(5), for clarity), but I'd rather focus on what you've done correctly with regard to the core of the work.

To be complete, you should include an explicit statement of the domain.

Now you need to do (a), which is a little harder in my opinion.

yes had mind blank currently correcting bare with