How do you find (f of g of h) given $f (x) = x^{2} + 1$ $f(x)=x^2+1$ $g (x) = 2 x$ $g(x)=2x$ and $h (x) = x - 1$ $h(x)=x-1$ ?

Question

2737 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-28T22:30:13

See solution below

ƒ(g(h(x)))

= ƒ(g(x - 1)
= ƒ(2(x - 1))
= ƒ(2x - 2)
= ƒ( $2 x - 2)^{2}$ $2x - 2)^2$ + 1
= $4 x^{2}$ $4x^2$ - 8x + 4 + 1
= $4 x^{2}$ $4x^2$ - 8x + 5

So, ƒ(g(h(x))) = $4 x^{2}$ $4x^2$ - 8x + 5

I hope I properly understood your question!

How do you find (f of g of h) given f(x)=x^2+1f(x)=x2+1f(x)=x^2+1 g(x)=2xg(x)=2xg(x)=2x and h(x)=x-1h(x)=x−1h(x)=x-1?