How do you find the inverse of #f(x) = 3x + 1# and is it a function?

Answer 1

See explanation...

Let #y = f(x) = 3x+1#
Subtract #1# from both ends to get:
#y - 1 = 3x#
Divide both sides by #3# and transpose to get:
#x = (y-1)/3#
Having expressed #x# in terms of #y#, we can deduce that:
#f^(-1)(y) = (y-1)/3#
Since #f^(-1)(y)# is uniquely defined for any #y in RR#, it is a function.
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Answer 2

To find the inverse of ( f(x) = 3x + 1 ), follow these steps:

  1. Replace ( f(x) ) with ( y ): ( y = 3x + 1 ).
  2. Swap the variables ( x ) and ( y ): ( x = 3y + 1 ).
  3. Solve this equation for ( y ): [ x = 3y + 1 ] [ x - 1 = 3y ] [ \frac{x - 1}{3} = y ]

So, the inverse function of ( f(x) = 3x + 1 ) is ( f^{-1}(x) = \frac{x - 1}{3} ).

Yes, the inverse function is a function because for every input ( x ) in the domain of ( f(x) ), there exists a unique output ( y ) in the domain of ( f^{-1}(x) ), and vice versa.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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