Let $f (x) = 1 - x$ $f(x)=1-x$ and $g (x) = x^{2}$ $g(x)= x^2$ and $h (x) = \frac{1}{x}$ $h(x)= 1/x$ , how do you find $f (g (x))$ $f(g(x))$ ?

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Answer 1 · 2018-08-14T16:17:48

$f (g (x)) = 1 - x^{4}$ $f(g(x))=1-x^4$

Here ,

$f (x) = 1 - x and g (x) = x^{2}$ $f(x)=1-x and g(x)=x^2$

Now ,

$f(g(x))=f(x^2)to[becauseg(x)=x^2]$

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

$f(g(x))=1-x^4$

Let f(x)=1-xf(x)=1−xf(x)=1-x and g(x)= x^2g(x)=x2g(x)= x^2 and h(x)= 1/xh(x)=1xh(x)= 1/x, how do you find f(g(x))f(g(x))f(g(x))?