How do you find $(f \circ g) (x)$ $(f @ g)(x)$ and its domain, $(g \circ f) (x)$ $(g @ f)(x)$ and its domain, $(f \circ g) (- 2)$ $(f @ g)(-2)$ and $(g \circ f) (- 2)$ $(g @ f)(-2)$ of the following problem $f (x) = x + 2$ $f(x) = x+ 2$ , $g (x) = 2 x^{2}$ $g(x) = 2x^2$ ?

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How do you find $(f \circ g) (x)$ $(f @ g)(x)$ and its domain, $(g \circ f) (x)$ $(g @ f)(x)$ and its domain, $(f \circ g) (- 2)$ $(f @ g)(-2)$ and $(g \circ f) (- 2)$ $(g @ f)(-2)$ of the following problem $f (x) = x + 2$ $f(x) = x+ 2$ , $g (x) = 2 x^{2}$ $g(x) = 2x^2$ ?

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Answer 1 · 2018-01-17T13:33:25

See the explanation below...

Explanation:

By the composition $(f \circ g) (x)$ $(f\circ g)(x)$ , we mean $f (g (x))$ $f(g(x))$ and by $(g \circ f) (x)$ $(g\circ f)(x)$ , we mean $g (f (x))$ $g(f(x))$ .

To find $f (g (x))$ $f(g(x))$ , we need to put the value of $g (x)$ $g(x)$ for every value of $x$ $x$ in $f (x)$ $f(x)$ . So, by doing this, we get:

$(f \circ g) (x) = f (g (x)) = (2 x^{2}) + 2$ $(f\circ g)(x)=f(g(x))=(2x^2)+2$

$= 2 x^{2} + 2$ $=2x^2+2$

The domain of a function is the set of values for which the function is real and defined.

This above evaluated function has no undefined points. The domain is $- \infty < x < \infty$ $-oo < x < oo$

Similarly, we have:

$(g \circ f) (x) = g (f (x)) = 2 {(x + 2)}^{2}$ $(g\circ f)(x)=g(f(x))=2(x+2)^2$

Simplify:

$= 2 x^{2} + 8 x + 8$ $=2x^2+8x+8$

Domain: $- \infty < x < \infty$ $-oo < x < oo$

Now, in the same manner, find $(f \circ g) (- 2)$ $(f\circ g)(-2)$ as:

$= 2 {(- 2)}^{2} + 2$ $=2(-2)^2+2$

Simplify:

$= 10$ $=10$

And:

$(g \circ f) (- 2) = 2 {(- 2)}^{2} + 8 (- 2) + 8$ $(g\circ f)(-2)=2(-2)^2+8(-2)+8$

Simplify:

$= 0$ $=0$

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How do you find $(f \circ g) (x)$ $(f @ g)(x)$ and its domain, $(g \circ f) (x)$ $(g @ f)(x)$ and its domain, $(f \circ g) (- 2)$ $(f @ g)(-2)$ and $(g \circ f) (- 2)$ $(g @ f)(-2)$ of the following problem $f (x) = x + 2$ $f(x) = x+ 2$ , $g (x) = 2 x^{2}$ $g(x) = 2x^2$ ?

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How do you find $(f \circ g) (x)$ $(f @ g)(x)$ and its domain, $(g \circ f) (x)$ $(g @ f)(x)$ and its domain, $(f \circ g) (- 2)$ $(f @ g)(-2)$ and $(g \circ f) (- 2)$ $(g @ f)(-2)$ of the following problem $f (x) = x + 2$ $f(x) = x+ 2$ , $g (x) = 2 x^{2}$ $g(x) = 2x^2$ ?