Question

A function f(x)=3x+12.

Answer 1

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•  Function:   f(x) = 3x + 12.


A.  Finding the inverse of f.

The composition of f with its inverse results in the identity function:

(f o g)(x) = x

f[ g(x) ] = x

3 · g(x) + 12 = x

3 · g(x) = x – 12

              x – 12
g(x)  =  ⸺⸺
                 3

               x 
g(x)  =  ⸺  –  4    <———    this is the inverse of f.
               3

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B.  Verifying that the composition of f and g gives us the identity function:

•  $\mathsf{(f\circ g)(x)}$

$\mathsf{=f\big[g(x)\big]}\\\\\\ \mathsf{=3\cdot \left(\dfrac{x}{3}-4\right)+12}\\\\\\ \mathsf{=\diagup\hspace{-7}3\cdot \dfrac{x}{\diagup\hspace{-7}3}-3\cdot 4+12}\\\\\\ \mathsf{=x-12+12}\\\\ \mathsf{=x\qquad\quad\checkmark}$


and also

•  $\mathsf{(g\circ f)(x)}$

$\mathsf{=g\big[f(x)\big]}\\\\\\ \mathsf{=\dfrac{f(x)}{3}-4}\\\\\\ \mathsf{=\dfrac{3x+12}{3}-4}\\\\\\ \mathsf{=\dfrac{\diagup\hspace{-7}3\cdot (x+4)}{\diagup\hspace{-7}3}-4}\\\\\\ \mathsf{=x+4-4}\\\\ \mathsf{=x\qquad\quad\checkmark}$

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C.  Since f and g are inverse, then

f(g(– 2))

= (f o g)(– 2)

= – 2          ✔


•  Call h the compositon of f and g. So,

h(x) = (f o g)(x)

h(x) = x


As you can see above, there is no restriction for h. Therefore, the domain of h is R (all real numbers).


I hope this helps. =)