Given the functions
f(x)=2x+1 ,g(x)=x+1 and h(x)=x^2
Find :(i) fogoh
So we have three functions, f, g, and h, which are defined in terms of the variable x.
Since it is not specified, presumably these functions have domain R.
The range can be discovered as follows:
if we stick a real number into f, it is scaled up by 2, and then 1 is added.
Given an arbitrary real number y, can we find an appropriate a such that f(a) yields y?
Yes. Choose a = (y-1)/2. Then f(a) would be:
f((y-1)/2) =2 ((y-1)/2)+1
Since y was arbitrary, and we could construct a, we see the range is all real numbers as well.
(I know this isn't your question yet... but we're getting there)
g(x)=x+1 also has range R, it is simply adding one unit to its input. I'll spare you the details.
Now, what about the next one, h?
h(x)=x^2 with domain R can never produce outputs less than zero.
Suppose we wanted -1 as output. Which real number can square to result in -1? There isn't one.
So in the case of the function h, the range will be [0,+infinity), i.e. all positive real numbers and zero.
Of course, if the domain is C, this information doesn't apply. Point being: domain and range matter.
Now, how do we compose these functions?
fogoh(x) = f(g(h(x))), and gohof(x) = g(h(f(x))) by definition of function composition, right?
Now we can start expanding these by using the function's specification as follows:
fogoh(x) = f(g(h(x))) definition of function composition
= f(g(x^2)) application of h
= f((x^2)+1) application of g
= 2((x^2)+1)+1 application of f
= 2(x^2)+2+1 distributive property
= 2(x^2)+3 simplified form
So fogoh(x) = 2(x^2)+3, and with domain R will have a range of [3,+infinity).
gohof(x) = g(h(f(x))) definition of function composition
= g(h(2x+1)) application of f
= g((2x+1)^2) application of h
= (2x+1)^2+1 application of g
= (2x)^2+2(2x)(1)+(1)^2+1 square of binomial
= 4(x^2)+4x+2 simplified form
Finally gohof(x) = 4(x^2)+4x+2, and with domain R will have a range of [1,+infinity).
How do we find the range in this case? The easy way would be to plot it, in which case
you'll see that with the input of -1/2, it takes on its minimum value, which is 1.
I hope this was helpful! Cheers with your revision.
Wow of course it was helpful:-) .Thanx
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