How do you find $(f@g)(x)$ given $g(x) = (2x) (1/2)$ , $f(x) = x^2 + 1$ ?

Question

4218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T16:40:10

See below.

$(f@g)(x)=f(g(x))=g(x)^2-1=x^2-1=f(x)$

Not weird,

because $g(x)=x$ is the identity transformation.

Answer 2 · 2016-10-21T17:04:39

$(f@g)(x) = f(x)$ .

Let's start by simplifying function $g$ .

$g(x) = 2x(1/2) = (2x)/2 = x$

Now, we can rewrite $(f @ g)(x)$ as $f(g(x))$ .

$f(g(x)) = (x)^2 + 1$

So, $f(g(x)) = f(x)$ .

Hopefully this helps!

How do you find (f@g)(x)(f@g)(x) given g(x) = (2x) (1/2)g(x) = (2x) (1/2), f(x) = x^2 + 1 f(x) = x^2 + 1 ?