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Let [tex]\( h(x) = (g \circ f)(x) = \frac{1}{(x+3)^2} \)[/tex]

Which of the following could be a possible decomposition of [tex]\( h(x) \)[/tex]?

A. [tex]\( f(x) = \frac{1}{x} \)[/tex]; [tex]\( g(x) = x+3 \)[/tex]

B. [tex]\( f(x) = \frac{1}{x^2} \)[/tex]; [tex]\( g(x) = x+3 \)[/tex]

C. [tex]\( f(x) = x^2 \)[/tex]; [tex]\( g(x) = x+3 \)[/tex]

D. [tex]\( f(x) = x+3 \)[/tex]; [tex]\( g(x) = \frac{1}{x^2} \)[/tex]


Sagot :

Consider the function [tex]\( h(x) = (g \circ f)(x) = \frac{1}{(x+3)^2} \)[/tex].

We need to find the correct decomposition of [tex]\( h(x) \)[/tex] into functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] such that [tex]\( h(x) = g(f(x)) \)[/tex].

Let's test each option one by one:

### Option A:
- [tex]\( f(x) = \frac{1}{x} \)[/tex]
- [tex]\( g(x) = x+3 \)[/tex]

Applying these:
[tex]\[ g(f(x)) = g\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right) + 3 \][/tex]
[tex]\[ g(f(x)) \neq \frac{1}{(x+3)^2} \][/tex]
Therefore, Option A is incorrect.

### Option B:
- [tex]\( f(x) = \frac{1}{x^2} \)[/tex]
- [tex]\( g(x) = x+3 \)[/tex]

Applying these:
[tex]\[ g(f(x)) = g\left(\frac{1}{x^2}\right) = \left(\frac{1}{x^2}\right) + 3 \][/tex]
[tex]\[ g(f(x)) \neq \frac{1}{(x+3)^2} \][/tex]
Therefore, Option B is incorrect.

### Option C:
- [tex]\( f(x) = x^2 \)[/tex]
- [tex]\( g(x) = x+3 \)[/tex]

Applying these:
[tex]\[ g(f(x)) = g\left(x^2\right) = \left(x^2\right) + 3 \][/tex]
[tex]\[ g(f(x)) \neq \frac{1}{(x+3)^2} \][/tex]
Therefore, Option C is incorrect.

### Option D:
- [tex]\( f(x) = x+3 \)[/tex]
- [tex]\( g(x) = \frac{1}{x^2} \)[/tex]

Applying these:
[tex]\[ g(f(x)) = g(x+3) = \frac{1}{(x+3)^2} \][/tex]
[tex]\[ g(f(x)) = \frac{1}{(x+3)^2} \][/tex]
This matches the given function [tex]\( h(x) \)[/tex].

Therefore, the correct possible decomposition of [tex]\( h(x) \)[/tex] is:
[tex]\[ \boxed{f(x) = x+3, \ g(x) = \frac{1}{x^2}} \][/tex]

The correct option is D.
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