Question

Question #9e398

Answer 1

Substitute for `f` and `g` in terms of `n` as described by the given equations, then simplify, to get:

`(f + g)(n/2) = ((2n - 2) + (3n^2 - 4))(n/2) = 3/2 n^3 + n^2 - 3n`

Explanation:

Let's look at what we have:

`f = 2n - 2`

`g = 3n^2 - 4`

And what we're "looking for":

`E = (f + g)(n/2)`

Where I set "E" for "expression".

If you're simply given a question this vague, I'd assume what is being asked is to rewrite the expression in terms of `n`. What that means is make sure `n` is the only variable, and that there is no `f` or `g`, but it still has to be "equivalent" (it has to mean the same thing) to the original expression.

In that case, simply substitute ("replace") each variable with the "fuller expression" as shown by the equations `f = 2n - 2` and `g = 3n^2 - 4`.

Starting with `f = 2n - 2`:

`E = (f + g)(n/2) = ((2n - 2) + g)(n/2)`

See what I did there? Since, by that equation, `f` is equal to `2n - 2`, that means I could, roughly speaking, write `2n - 2` anywhere and it would automatically mean `f`, and vise versa, I could write `f` anywhere and it would automatically mean `2n - 2`.

Now let's do the same for `g = 3n^2 - 4`:

`E = ((2n - 2) + g)(n/2) = ((2n - 2) + (3n^2 - 4))(n/2)`

Alright, now we have successfully rewritten the given expression in terms of `n`, there are no `f`s or `g`s left there, but... it looks like we can do some simplifying ("cleaning-up").

`E = ((2n - 2) + (3n^2 - 4))(n/2) = (3n^2 + 2n - 6)(n/2)`

What I just did is "combine like terms": since we have `-2` and `-4`, that just combines to `-6`. I also ordered it in the conventional way polynomials are written: start from the highest "degree" (exponent).

But don't forget to multiply `n/2` over:

`E = (3n^2 + 2n - 6)(n/2) = (3n^3 + 2n^2 - 6n)/2`

By the distributive property, we could multiply an entire sum by multiplying each term, each "number/thing" inside it that is being added up. Then, we have division by 2, so why not divide the entire thing?

We could also apply that same distirbutive property for division:

`E = (3n^3 + 2n^2 - 6n)/2 = 3/2 n^3 + n^2 - 3n`

Some terms may seem to have "lost" some amount of their coefficients, because it has been divided by 2.

I think this looks like a final answer.

So, to conclude, given `f = 2n - 2` and `g = 3n^2 - 4`, we have that `(f + g)(n/2)` simplifies to `3/2 n^3 + n^2 - 3n`.

Answer 2

`1.5n^3+n^2-3n`

Explanation:

We have:

`(f+g)*(n/2)`

Let's rewrite this as:

`(f+g)*n*1/2`

Let's replace `f` and `g` in terms of `n`.

=>`(2n-2+(3n^2-4))*n*1/2` distribute the positive sign (which changes nothing.)

=>`(2n-2+3n^2-4)*n*1/2` Reorder the expression so that the terms with higher degree (exponent) comes first.

=>`(3n^2+2n-2-4)*n*1/2` Combine like terms

=>`(3n^2+2n-6)*n*1/2` Distribute `n`. (multiply all terms in the parenthesis by `n`)

=>`(3n^2*n+2n*n-6*n)*1/2`

=>`(3n^3+2n^2-6n)*1/2`

=>`(3n^3+2n^2-6n)*1/2` Distribute 1/2 just like we did with `n`.

=>`3n^3*1/2+2n^2*1/2-6n*1/2` Multiplication is commutative!

=>`3*1/2n^3+2*1/2n^2-6*1/2n`

=>`3/2n^3+2/2n^2-6/2n` You can turn the fractions to integers or decimals.

=>`1.5n^3+1n^2-3n`

=>`1.5n^3+n^2-3n`

Answer 3

`(f+g)(n/2) = f(n/2)+g(n/2) = 3/4n^2+n-6`

please see disclaimer below, however

Explanation:

Disclaimer
I am assuming that the question asks for the value the function `f+g` takes when it is given the argument `n/2` (this is not very clear from the typography - which is why the other two answers provided are different from this one).

`f(n) = 2n-2, qquad g(n) = 3n^2-4`

So,
`f(n/2) = 2times n/2-2 = n-2`
and
`g(n/2) = 3 times (n/2)^2-4 = 3/4n^2-4`
so that

`(f+g)(n/2) = f(n/2)+g(n/2) = 3/4n^2+n-6`

Answer 4

Given
`f(n)=2n-2`
`g(n)=3n^2-4`

`:.(f+g)(n)=((2n-2)+(3n^2-4))`
`=>(f+g)(n)=2n+3n^2-6` .......(1)

We are to find `(f+g)(n/2)`
Changing `nto n/2` in (1) we get

`(f+g)(n/2)=2n/2+3(n/2)^2-6`
`=>(f+g)(n/2)=n+3/4n^2-6`
`=>(f+g)(n/2)=3/4n^2+n-6`