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Marin writes the functions m(x) = startfraction 3 x over x + 7 endfraction and n(x) = startfraction 7 x over 3 minus x endfraction. Which equation must be true for m(x) and n(x) to be inverse functions?.

Sagot :

Equation which represents the m(x) and n(x) are inverse functions given by Option B.

3[7x/(3 - x)] / [ ( 7x / 3 - x ) + 7 ] =7[3x/(x + 7)] / [ 3 - ( 3x / x + 7 )  ] =x.

As given in the question,

Given functions are

m(x) = 3x /(x+7) and n(x) = 7x/ (3 - x)

Let y = m(x)

y = 3x /(x+7)

Now replace x by y and vice versa:

x = 3y / (y + 7) __(1)

Now substitute (1) in n(x)

n(x) = 7x/ (3 - x)

      = 7[3y / (y + 7)] / [ 3 - 3y / (y + 7) ]  __(3)

      = y

Now n(x) = y

y = 7x/ (3 - x)

Replace x by y and vice versa

x = 7y /( 3 - y )  __(2)

Now , substitute (2) in m(x)

m(x) = 3x /(x+7)

       = 3[7y /( 3 - y ) ] / [ 7y /( 3 - y )  + 7 ]  __(4)

       = y

Replace y by x in (3) and (4)

Equation representing m(x) and n(x)  as inverse functions:

3[7x/(3 - x)] / [ ( 7x / 3 - x ) + 7 ] =7[3x/(x + 7)] / [ 3 - ( 3x / x + 7 )  ] =x.

Therefore, equation representing m(x) and n(x) as inverse function is given by :

Option B. 3[7x/(3 - x)] / [( 7x / 3 - x ) + 7] =7[3x/(x + 7)] /[3 - ( 3x / x + 7 )] =x.

   

The complete question is :

Marin write the functions m(x) = 3x /(x+7) and n(x) = 7x/ (3 - x).

Which equation must be true for m(x) and n(x) to be inverse functions?.

Options are shown in attachment.

Learn more about functions here

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