Algebra functions Core 3 maths

[sa1994]

How do I work out gh(x) and give the domain and range of this? When g(x)= root x where x>_0 and h(x) 2x+1 where xE

I have started by doing g(2x+1)= root x(2x+1)

but I dont know where to go from here? I would appreciate some help.

 

[Deleted member 4993]

How do I work out gh(x) and give the domain and range of this? When g(x)= root x where x>_0 and h(x) 2x+1 where xE

I have started by doing g(2x+1)= root x(2x+1)

but I dont know where to go from here? I would appreciate some help.

What does root mean? square -root? cube-root?? "root of a function"??? which one????

 

[HallsofIvy]

How do I work out gh(x) and give the domain and range of this? When g(x)= root x where x>_0 and h(x) 2x+1 where xE

Isn't there something missing here?

I have started by doing g(2x+1)= root x(2x+1)

but I dont know where to go from here? I would appreciate some help.

Assuming you mean square root and that \(\displaystyle x\in R\), the the square root can be applied only to non-negative numbers so we must have \(\displaystyle 2x+ 1\ge 0\). Solve that to get the domain. As for range, since \(\displaystyle g(x)= \sqrt{x}\) is define as "the positive number, y, such that \(\displaystyle y^2= x\)", gh clearly can't be negative. Can it ever be 0?

It also is NOT a good idea to write gh(x) as "root x(2x+1)". Even assuming that "root" means square root, that means \(\displaystyle g(2x+1)= \sqrt{x(2x+1)}\) which is not correct.

 

[JeffM]

How do I work out gh(x) and give the domain and range of this? When g(x)= root x where x>_0 and h(x) 2x+1 where xE

I have started by doing g(2x+1)= root x(2x+1)

but I dont know where to go from here? I would appreciate some help.

We appreciate that you have shown what you have thought about, but please try to make your questions complete and clear.

As Subhotosh Khan said, root is ambiguous. You can use fractional exponents such as g(x) = x^(1/2) to express roots clearly.

No one has a clue what xE is supposed to mean.

Is this the problem: \(\displaystyle Given\ g(x) = \sqrt{x}\ and\ h(x) = 2x + 1,\ find\ f(x) = g(h(x)),\ its\ domain,\ and\ its\ range.\)

Note that using my notation you work from the inside out. For g(h(x)), render h(x) as you normally would, and use that result as the input to the function g.

So \(\displaystyle f(x) = g(h(x)) = g(2x + 1) = \sqrt{2x + 1} \ne \sqrt{x(2x + 1)}.\)

This is the key to understanding composite functions. The output of the innermost function is used as the input to the next function. Are you sure you have this?

Now that you have f(x), you figure out the domain and range the regular way.