Does anyone understand this? Find [f ○ (g ○ h)](2) if f(x) = 2x - 1, g(x) = 4x, and h(x) = x2+ 1. ??

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Answer 1 · 2018-04-03T17:49:51

$39$ $color(blue)(39$

$f \circ g \circ h \Leftrightarrow f (g (h (x)))$ $f@g@h<=>f(g(h(x)))$

Working from the inside out.

$h (x) = x^{2} + 1$ $h(x)=x^2+1$ , $g (x) = 4 x$ $g(x)=4x$

$g (h (x)) = 4 (h (x)) = 4 (x^{2} + 1)$ $g(h(x))=4(h(x))=4(x^2+1)$

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

$f (g (h (x)) = 2 (g (h (x))) - 1 = 2 (4 (x^{2} + 1)) - 1$ $f(g(h(x))=2(g(h(x)))-1=2(4(x^2+1))-1$

$= 2 (4 x^{2} + 4) - 1 = 8 x^{2} + 8 - 1 = 8 x^{2} + 7$ $=2(4x^2+4)-1=8x^2+8-1=color(blue)(8x^2+7)$

$f (g (h (x))) (2) = 8 {(2)}^{2} + 7 = 39$ $f(g(h(x)))(2)=8(2)^2+7=color(blue)(39$

Answer 2 · 2018-04-04T13:30:47

For a different approach, see below.

$[f \circ (g \circ h)] (2) = f ([g \circ h] (2))$ $[f@(g@h)] (2) = f([g@h] (2))$

$= f (g (h (2))$ $= f(g(h(2))$

We'll start by finding $h$ $h$ of $2$ $2$ .

We have $h (x) = x^{2} + 1$ $h(x) = x^2+1$ , so $h (2) = 2^{2} + 1 = 5$ $h(2) = 2^2+1=5$ .

Therefore,

$f (g (h (2)) = f (g (5))$ $f(g(color(red)(h(2))) = f(g(color(red)(5)))$ .

Now find $g (h (2))$ $g(h(2))$ which is the same as $g$ $g$ of $5$ $5$ ,

Since $g (x) = 4 x$ $g(x) = 4x$ , we get $g (5) = 4 (5) = 20$ $g(5)=4(5)=20$ .

So,

$f (g (5)) = f (20)$ $f(g(5)) = f(20)$ .

Finally, we'll find $f$ $f$ of $20$ $20$ .

Since $f (x) = 2 x - 1$ $f(x) = 2x-1$ , we finish with

$f (20) = 2 (20) - 1 = 39$ $f(20) = 2(20)-1 = 39$