What is the inverse of the function f(x) = 2x + 1?
1
o h(x) = 5x-
2x +
O h(x) = 1
3x-2
N/" Nr
Oh(x) =
O h(x) = x + 2

Respuesta :

Answer:

[tex]f^{-1}[/tex] (x) = [tex]\frac{x-1}{2}[/tex]

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = 2x + 1 ( subtract 1 from both sides )

y - 1 = 2x ( divide both sides by 2 )

[tex]\frac{y-1}{2}[/tex] = x

Change y back into terms of x with x = [tex]f^{-1}[/tex] (x) , so

[tex]f^{-1}[/tex] (x) = [tex]\frac{x-1}{2}[/tex]

Answer:  " f⁻¹(x) = (x-1)/2  ";  or, write as:

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              " f⁻¹(x) = (x/2) - (1/2) " .

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Step-by-step explanation:

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Given:   f(x) = 2x + 1 ;  Find the inverse:

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1)  Rewrite as y = 2x + 1 ;

2) Replace the "x" with "y" ; and the "y" with "x" ; and rewrite:

         x = 2y + 1 ;

3)  Now, "solve"; with "y" standing alone as a single, isolated variable on the left side of the equation, with an "equals" sign following the "y" :

         x = 2y + 1 :

Subtract "1" from each side of the equation:

         x - 1 = 2y + 1 - 1 ;

to get:

         x - 1 = 2y ;

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↔  Rewrite as :

2y = x - 1 ;

Now, divide each side of the equation by "2" ;

  to isolate "y" on the left-hand side of the equation;

  & to solve in terms of "y":

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2y / 2 = (x-1)/2 ;

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to get:

y = (x-1)/2 ;

or; write as:  y = (x/2) - (1/2);

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Now, rewrite; by replacing the "y" with "f⁻¹(x) "; as follows (to indicate that this is the "inverse function"):

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f⁻¹(x) = (x-1)/2 ;  or, write as:

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f⁻¹(x) = (x/2) - (1/2) .

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Hope this is helpful to you!

 Best wishes!

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