Let $f (x) = \frac{1}{x - 2}$ and $g (x) = \frac{2}{x} + 2$ .
Then $(f \circ g) (x) =$
$(g \circ f) (x) =$

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Answer

Therefore, we have $(f \circ g) (x) = \frac{x}{2}$ and $(g \circ f) (x) = 2 x - 2$ .

Steps

Step 1 ：Let $f (x) = \frac{1}{x - 2}$ and $g (x) = \frac{2}{x} + 2$ .

Step 2 ：The composition of two functions, $f$ and $g$ , denoted as $(f \circ g) (x)$ or $f (g (x))$ , is a function that applies $g$ to its input, and then $f$ to the result.

Step 3 ：Similarly, $(g \circ f) (x)$ or $g (f (x))$ is a function that applies $f$ to its input, and then $g$ to the result.

Step 4 ：To find $(f \circ g) (x)$ , we need to substitute $g (x)$ into $f (x)$ , and to find $(g \circ f) (x)$ , we need to substitute $f (x)$ into $g (x)$ .

Step 5 ：The composition of the functions $f$ and $g$ is given by $(f \circ g) (x) = \frac{x}{2}$ and $(g \circ f) (x) = 2 x - 2$ .

Step 6 ：Therefore, we have $(f \circ g) (x) = \frac{x}{2}$ and $(g \circ f) (x) = 2 x - 2$ .

Let f(x)=1x−2 and g(x)=2x+2.Then (f∘g)(x)=(g∘f)(x)=

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Popular Problems

Let $f (x) = \frac{1}{x - 2}$ and $g (x) = \frac{2}{x} + 2$ .
Then $(f \circ g) (x) =$
$(g \circ f) (x) =$