HI I need help with composition in math.?

Question

here is the problem,..........................................
f(x)=3x+5 g(x)=2x+4 h(x)=6-x^
-------
x-2
g(x) is a fraction

now here is the problem
f(x^)-2h(x)=
the equation is f(x squared I hope I did it properly)
Please can you explain this in steps?? Thank you.

here is the qestion for g(x)=2x=4
____
x-2

here is the qestion for g(x)=2x=4/x-2
it is supposed to read 2x-4 over x-2

Answer 1

you do not need g(x) for this question.
you only need f(x) and h(x)
also, x squared is written as x^2

so take the x in f(x) and square it, and take the whole h(x) and times it by 2, you get,

3x^2+5 - 12+2x^2
3x^2+2x^2 is 5x^2, 5-12 is -7
therefore you get
b(x) = 5x^2 -7
if you equate it to 0 to find x, you get
5x^2 = 7
x^2 = 7/5
x = +/- sqrt (7/5)

side note: the reason why f(x^2) is 3x^2+5 is because it is only the x that gets squared within the function. if it is (f(x))^2, then it will be the entire function, (3x+5)^2

Answer 2

I'm assuming the carets are indicating a variable is squared.
1. The x^2 in the problem indicates that x^2 should be substituted in the x-dependent side of the f of x function.
3(x^2)+5-2h(x)
2. h(x) can now be substituted for also.
3x^2 + 5 - 2(6-x^2)
3. Now distribute.
3x^2 + 5 - 12 + 2x^2
4. Add like terms.
5x^2 - 7

Answer 3

Hi there =)

I don't understand how g(x) is a fraction, unless the x value you give it is a fraction that won't reduce away, since g(1/2) = 5, but g(1/3) = 14/3

In your question, twice you typed "x^" by itself, so I am going to assume you meant x squared both times since you mentioned that in your post. x squared would be shown as x^2. If that's not what you meant both times, please re-interpret my answer in the proper way.

So our h(x) = 6-x^2

If I understand, you are looking to create a new function like this:

j(x) = f(x^2) - 2h(x)

For our f(x) funtion, let's replace all of our x's with x^2 and plug them into our function:

j(x) = 3x^2 + 5 -2h(x)

Now let's replace h(x) with it's actually value, but we have to be careful, because 2 is multiplying the whole function, not just one term of it.

j(x) = 3x^2 + 5 -2(6 - x^2)
j(x) = 3x^2 + 5 -12 + 2x^2
j(x) = 5x^2 - 7

That is our new function, which equals f(x^2) - 2h(x)

--charlie