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Mathematics 20 Online
OpenStudy (anonymous):

I am trying to find the domain of composite function f(g(x). f(x)=x^2+1 and g(x)=square root of x-1.

ganeshie8 (ganeshie8):

domain of a composite function is the intersection of domains of its parent funcitons

ganeshie8 (ganeshie8):

domain of f(g(x)) = domain of f(x) \(\cap \) domain of g(x)

OpenStudy (anonymous):

What would happen if I plug in g(x) into f(x)?

ganeshie8 (ganeshie8):

plugin and see

ganeshie8 (ganeshie8):

\(f( \color{red}{x} )=\color{red}{x}^2+1\) \(f( \color{red}{g(x)} ) = (\color{red}{g(x)})^2 + 1 \)

OpenStudy (anonymous):

then it is square root of x-1 to the second power plus one, right?

ganeshie8 (ganeshie8):

yup !

OpenStudy (anonymous):

then, how do I find the domain?

ganeshie8 (ganeshie8):

\(f( \color{red}{x} )=\color{red}{x}^2+1\) \(f( \color{red}{g(x)} ) = (\color{red}{g(x)})^2 + 1 \) \(\qquad ~~= (\sqrt{x-1})^2 + 1 \)

ganeshie8 (ganeshie8):

since f(g(x)) is a composition of f(x) and g(x), it needs to SATISFY BOTH the domains of f(x) and g(x)

ganeshie8 (ganeshie8):

can u find domain of f(x) and g(x) ?

OpenStudy (anonymous):

Let's see .. the domain of f(x) is all real numbers and the domain of g(x) is x is greater of equal to one.

OpenStudy (anonymous):

I mean than greater of equal to one

ganeshie8 (ganeshie8):

yes, domain of f(x) : all reals domain of g(x) : \( x\ge 1\)

ganeshie8 (ganeshie8):

so, the domain of \(f(g(x))\) is \(x \ge 1\) cuz \(x \ge 1\) is the intersection of both domains

ganeshie8 (ganeshie8):

another way to look at it is : \(x \ge 1\) satisfies both given functions' domains f(x) and g(x)

OpenStudy (anonymous):

ok thank you

ganeshie8 (ganeshie8):

u wlc :)

ganeshie8 (ganeshie8):

when u have time watch this video : https://www.youtube.com/watch?v=_zy7Uro7iCg

ganeshie8 (ganeshie8):

actually i was wrong earlier, to find domain of f(g(x)) u need to find intersection of domains of g(x) and f(g(x)) however the answer is still same here : x >= 1 is the domain of f(g(x))

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